Fluid Mechanics for Master Students

Transcription

1 Fluid Mechanics for Master Students Olivier Cleynen A one-semester course for students in the Chemical and Energy Engineering program at the Otto-von-Guericke-Universität Magdeburg last edited May 4,

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3 Contents Syllabus 5 0 Important concepts Concept of a fluid Fluid mechanics Important concepts in mechanics Properties of fluids Forces on fluids Basic flow quantities Five equations Classification of fluid flows Limits of fluid mechanics Exercises Effects of pressure Motivation Concept of pressure Pressure in static fluids Wall pressure forces and buoyancy Exercises Effects of shear Motivation Concept of shear Slip and viscosity Wall shear forces Exercises Analysis of existing flows Motivation The Reynolds transport theorem Mass conservation Change of linear momentum Change of angular momentum Energy conservation The Bernoulli equation Limits of integral analysis Exercises Predicting fluid flow Motivation Eulerian description of fluid flow Mass conservation Change of linear momentum Change of angular momentum Energy conservation and increase in entropy CFD: the Navier-Stokes equations in practice Exercises

4 5 Pipe flow Motivation Perfect flow in ducts Viscous laminar flow in ducts Viscous laminar flow in cylindrical pipes Viscous turbulent flow in cylindrical pipes Exercises Scale effects Motivation Scaling forces Scaling flows Flow parameters as force ratios The Reynolds number in practice Exercises Flow near walls Motivation Describing the boundary layer The laminar boundary layer Transition The turbulent boundary layer Separation Exercises Modeling large- and small-scale flows Motivation Flow at large scales Plotting velocity with functions Flow at very small scales Exercises Introduction to compressible flow Motivation Compressibility and its consequences Thermodynamics of isentropic flow for a perfect gas Speed and cross-sectional area Isentropic flow in converging and diverging nozzles The perpendicular shock wave Compressible flow beyond frictionless pipe flow Exercises Appendix 215 A1 Notation A2 Field operators A3 List of references

5 Syllabus Fluid Mechanics for Master Students Summer semester 2018 Olivier Cleynen Welcome to the Fluid Mechanics course of the Chemical and Energy Engineering program! Objectives This one-semester course is designed for students working towards a Master in engineering, who have had little or no previous experience with fluid dynamics. My objectives are simultaneously: to provide you with a solid understanding of what can, an cannot, be done with fluid mechanics in engineering; to enable you to solve selected engineering fluid dynamics problems with confidence; to request the minimum feasible amount of your time and energy for the above. If all goes well, at the end of the semester, you should be well-prepared to begin a course in Computational Fluid Dynamics, where the knowledge and skills you acquire here can be used to solve applied problems in great detail. Course homepage The course homepage is located at a hard-to-unremember location: From that page, you can download the latest version of the present document, as well as lecture slides and archived examinations. The course script is the only document you need to work through and succeed in this course. I highly recommend that you print it out (at 200 pages, a copy at the University print shop costs less than 9 ). If you can t afford this, print at least the exercise sheets and the first page of each chapter. Lectures We will have lectures on Thursdays from 11:15 to 12:45 in room G05-H4. In class, we will cover a lot of content, at a rather brisk pace. The most efficient way to proceed is to read the notes before the lecture. Try doing that: even a brief skim over the relevant chapter helps. We will also experiment with new tools: if you have one at hand, try bringing an Internet-connected device with you. 5

6 Exercise sessions Exercise sessions take place on Fridays in room G22A-209. The schedule is slightly more complicated: For first-semester students, on odd weeks, from 13:00 to 15:00, and on even weeks, from 15:00 to 17:00; For second (or later) semester students, just the opposite: on even weeks, from 13:00 to 15:00, and on odd weeks, from 15:00 to 17:00. I am available during those sessions to help you work through the exercises, and will strive to answer all of your questions. The objective should be for you to work as fast as possible in order to cover each sheet entirely; attempting the first exercises on the prior day helps greatly in this respect. Although I occasionally use the board and projector, these sessions are hands-on and you should expect to work autonomously, alone or in groups. Bring your own exercise sheets, a calculator, a sugary drink, and start right away! Assessment The end-of-semester examination will take place on July 12 th, from 15:00 to 17:00 in room G26-HS1. This examination is the only assessment for the course. It lasts exactly 2 hours and is closed-book (a formula sheet is provided, containing the formulas and data contained in the preamble of every exercise sheet). A briefing regarding the examination will take place in class immediately after we cover Chapter 5, on May 24 th. The examination consists exclusively of lightly-modified exercise sheet problems. As reported by students in the previous years, the exam is very hard if you haven t done the exercises, and very easy if you have done them. It rewards your ability to methodically, resolutely and reliably solve the problems studied in class. To succeed, make sure you can solve all of them without outside help! A handful of archived examinations, together with their full solutions, is provided on the course homepage to help you practice. Books The present document is, I hope, enough to allow you to work and succeed through the entire course. If you need a second source, I recommend the books of Crowe et al. [10] (particularly since a dozen copies are available in our library), White [13], Çengel et al. [16], or Munson et al. [18] (a few personal copies of which can be borrowed from my office). These are solid, beautifully-illustrated, well-established textbooks, with plenty of worked-out exercises. You will see that this course matches closely their content and style. 6

7 Time plan Our time plan, which is affected by the holidays in Saxony-Anhalt, should be as follows: Week 14: Ch. 0 (Important concepts) Week 15: Ch. 1 (Effects of pressure), No exercise session Week 16: Ch. 2 (Effects of shear) Week 17: Ch. 3 (Analysis of existing flows) part 1/2 Week 18: Ch. 3 (Analysis of existing flows) part 2/2 Week 19: No lecture Week 20: Ch. 4 (Predicting fluid flow) Week 21: Ch. 5 (Flow in pipes), Exam briefing Week 22: No lecture Week 23: Ch. 6 (Scale effects) Week 24: Ch. 7 (Flow near walls) Week 25: Ch. 8 (Large- and small-scale flows), Feedback Week 26: Ch. 9 (Introduction to compressible flow) Week 27: revision & practice Week 28: Examination: July 12 th, from 15:00 to 17:00 in room G26-HS1 Copyright and remixing This document still contains several figures from cited, fully-copyrighted works. The source files for all of the remaining figures (which stem from many different authors, not all associated with this course) can be accessed and re-used by clicking the links in the captions. The text of this document is licensed under a Creative Commons CC-by-sa (attribution, share-alike) license. The Latex sources can be downloaded from the git repository accessed from the course homepage. I am grateful for receiving all types of feedback, including reports of inaccuracies and errors. Contact I can always be reached by (olivier.cleynen@ovgu.de). It is far more efficient to ask questions during exercise sessions. If you cannot make it there or for urgent matters, contact me directly by mail (please be precise and succinct) or afternoons at my office G It s a pleasure to join you for this course this semester! Fluid mechanics is one of the most exciting disciplines out there. Now, let us begin! Olivier Cleynen April

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9 Fluid Mechanics Chapter 0 Important concepts last edited April 2, Concept of a fluid Fluid mechanics Solution of a flow Modeling of fluids Theory, numerics, and experiment Important concepts in mechanics Position, velocity, acceleration Forces and moments 12 Forces and moments Energy Properties of fluids Density Phase Temperature Perfect gas model Speed of sound Viscosity Forces on fluids Gravity Pressure Shear Basic flow quantities Five equations Classification of fluid flows Limits of fluid mechanics Exercises 23 These lecture notes are based on textbooks by White [13], Çengel & al.[16], and Munson & al.[18]. 0.1 Concept of a fluid We call fluid a type of matter which is continuously deformable, and which spontaneously tends to adapt its shape to its container by occupying all of the space made available to it. 0.2 Fluid mechanics Solution of a flow Fluid mechanics is the study of fluids subjected to given constraints. The most common type of problem in this discipline is the search for a complete description of the fluid flow around or through a solid object. This problem is solved when the entire set of velocities of fluid particles has been described. This set of velocities, which is often a function of time, can be described either as a set of discrete values ( pixelized data) or as a mathematical function; it is called the flow solution. 9

10 If the solution is known, the shear and pressure efforts generated on the surface of the object can be calculated. Other quantities, such as the force and moments applying on the object, or the fluid s energy gains or losses, can also be calculated Modeling of fluids Like all matter, fluids are made of discrete, solid molecules. However, within the scope of fluid mechanics we work at the macroscopic scale, in which matter can be treated like a continuum within which all physical properties of interest can be continuously differentiable. There are about molecules in the air within an empty 1-liter bottle at ambient temperature and pressure. Even when the air within the bottle is completely still, these molecules are constantly colliding with each other and with the bottle walls; on average, their speed is equal to the speed of sound: approximately km/h. Despite the complexity of individual molecule movements, even the most turbulent flows can be accurately described and solved by considering the velocities of groups of several millions of molecules collectively, which we name fluid particles. By doing so, we never find out the velocity of individual molecules: instead, those are averaged in space and time and result in much simpler and smoother trajectories, which are those we can observe with macroscopic instruments such as video cameras and pressure probes. Our selection of an appropriate fluid particle size (in effect defining the lower boundary of the macroscopic scale), is illustrated in fig We choose to reduce our volume of study to the smallest possible size before the effect of individual molecules becomes meaningful. Adopting this point of view, which is named the continuum abstraction, is not a trivial decision, because the physical laws which determine the behavior of molecules are very different from those which determine the behavior of elements of fluid. For example, in fluid mechanics we never consider any inter-element attraction or repulsion forces; while new forces appear due to pressure or shear effects that do not exist at a molecular level. A direct benefit of the continuum abstraction is that the mathematical complexity of our problems is greatly simplified. Finding the solution for the Figure 0.1 Measurement of the average value of a property (here, velocity V ; but it could be pressure, or temperature) inside a given volume. As the volume shrinks towards zero, the fluid can no longer be treated as a continuum; and property measurements will oscillate wildly. Figure CC-by-sa o.c. 10

11 bottle of still air mentioned above, for example, requires only a single equation (section p.33) instead of a system of equations with unknowns (all leading up to V average,x,y,z,t = 0!). Another consequence is that we cannot treat a fluid as if it were a mere set of marbles with no interaction which would hit objects as they move by. Instead we must think of a fluid even a low-density fluid such as atmospheric air as an infinitely-flexible medium able to fill in almost instantly all of the space made available to it Theory, numerics, and experiment Today, fluid dynamicists are typically specializing in any one of three subdisciplines: Analytical fluid mechanics which is the main focus of these lectures and which consists in predicting fluid flows mathematically. As we shall see, it is only able to provide (exact) solutions for very simple flows. In fluid mechanics, theory nevertheless allows us to understand the mechanisms of complex fluid phenomena, describe scale effects, and predict forces associated with given fluid flows; Numerical fluid mechanics also called Computational Fluid Dynamics or cfd, which consists in solving problems using very large numbers of discrete values. Initiated as a research topic in the 1970s, cfd is now omnipresent in the industry; it allows for excellent visualization and parametric studies of very complex flows. Nevertheless, computational solutions obtained within practical time frames are inherently approximate: they need to be challenged using analysis, and calibrated using experimental measurements; Experimental fluid mechanics which consists in reproducing phenomena of interest within laboratory conditions and observing them using experimental techniques. A very mature branch (it first provided useful results at the end of the 19 th century), it is unfortunately associated with high human, equipment and financial costs. Experimental measurements are indispensable for the validation of computational simulations; meanwhile, the design of meaningful experiments necessitates a good understanding of scale effects. Our study of analytical fluid mechanics should therefore be a useful tool to approach the other two sub-disciplines of fluid mechanics. 0.3 Important concepts in mechanics Mechanics in general deals with the study of forces and motion of bodies. A few concepts relevant for us are recalled here Position, velocity, acceleration The description of the movement of bodies (without reference to the causes and effects of that movement) is called kinematics. 11

12 The position in space of an object can be fully expressed using three components (one for each dimension) This is usually done with a vector (here written x). If the object moves, then this vector varies with time. The velocity V of the object is the rate of change in time of its position: V d x dt (0/1) The magnitude V of velocity V is called speed and measured in m s 1. Contrary to speed, in order to express velocity completely, three distinct values (also each in m s 1 ) must be expressed. Generally, this is done by using one of the following notations: V = u v w = u x u y u z = (u i ) (u 1,u 2,u 3 ) = u r u θ u z The acceleration a of the object is the rate of change in time of its velocity: a d V dt (0/2) Acceleration is especially important in mechanics because it can be deduced from Newton s second law (see eq. 0/23 p.19 below) if the forces applying on the object are known. Acceleration can then be integrated with respect to time to obtain velocity, which can be integrated with respect to time to obtain position. Like velocity, acceleration has three components (each measured in m s 2 ). It shows at which rate each component of velocity is changing. It may not always point in the same direction as velocity (fig. 0.2). 12

13 Figure 0.2 Acceleration vectors of a body following a curved trajectory when its speed (the magnitude of its velocity) remains constant (top) and when it changes continuously (bottom) Figure CC-0 o.c Forces and moments A force expresses an effort exerted on a body. Force has a direction as well as a magnitude, and so it is expressed with a vector (typically noted F). The three main types of forces relevant to fluid mechanics are those due to pressure, those due to shear, and those due to gravity (the force due to gravity is called weight). When a force exerts at a position r away from a reference point, it exerts a torsion (or twisting ) effort named moment. Like a force, a moment has a direction as well as a magnitude, and so is best expressed with a vector. This vector M is expressed as the cross-product of the arm r and the force F: M r F (0/3) Energy Energy, measured in joules (J), is in most general terms the ability of a body to set other bodies in motion. It can be accumulated or spent by bodies in a large number of different ways. The most relevant forms of energy in fluid mechanics are: Kinetic energy E k, accumulated as motion E k 1 2 mv 2 (0/4) Work W, which is energy spent on displacing an object over a distance l with a force F: W F l (0/5) 13

14 Internal energy I stored as heat within the body itself. As long as no phase changes occurs, the internal energy I of fluids is roughly proportional to their absolute temperature T. 0.4 Properties of fluids Beyond velocity, which is the primary unknown for us in fluid mechanics, there are a few other important fluid properties Density The density ρ is defined as the inverse of mass-specific volume v, ρ 1 v = m V (0/6) where ρ the density (kg m 3 ); v is the specific volume (m 3 kg 1 ); m is the considered mass (kg); and V is the considered volume (m 3 ). At ambient conditions air has a density of approximately ρ air = 1,2 kg m 3 ; that of water is almost a thousand times greater, at ρ water = kg m Phase Fluids can be broadly classified into phases, which are loosely-defined sets of physical behaviors. Typically one distinguishes liquids which are fluids with large densities on which surface tension effects play an important role, from gases or vapors which have low densities and no surface tension effects. Phase changes are often brutal (but under specific conditions can be blurred or smeared-out); they usually involve large energy transfers. The presence of multiple phases in a flow is an added layer of complexity in the description of fluid phenomena Temperature Temperature is a scalar property measured in Kelvins (an absolute scale). It represents a body s potential for receiving or providing heat and is defined, in thermodynamics, based on the transformation of heat and work. A conversion between Kelvins and degrees Celsius by subtracting 273,15 units: T ( C) = T (K) 273,15 (0/7) 14 Although we can feel temperature in daily life, it must be noted that the human body is a very poor thermometer in practice. This is because we constantly produce heat, and we infer temperature by the power our body loses or gains as heat. This power not only depends on our own body temperature (hot water feels hotter when we are cold), but also on the heat capacity of the fluid (cold water feels colder than air at the same temperature) and on the amount of convection taking place (ambient air feels colder on a windy day).

15 In spite of these impressions, the fact is that the heat capacity of fluids is extremely high (c air 1 kj kg 1 K, c water 4 kj kg 1 K). Unless very high velocities are attained, the temperature changes associated with fluid flow are much too small to be measurable in practice Perfect gas model Under a specific set of conditions (most particularly at high temperature and low pressure), several properties of a gas can be related easily to one another. Their absolute temperature T is then modeled as a function of their pressure p with a single, approximately constant parameter R pv/t : pv = RT (0/8) p = RT (0/9) ρ where R depends on the state and nature of the gas (J K 1 kg 1 ); and p is the pressure (Pa), defined later on. This type of model (relating temperature to pressure and density) is called an equation of state. Where R remains constant, the fluid is said to behave as a perfect gas. The properties of air can satisfactorily be predicted using this model. Other models exist which predict the properties of gases over larger property ranges, at the cost of increased mathematical complexity. Many fluids, especially liquids, do not follow this equation and their temperature must be determined in another way, most often with the help of laboratory measurements Speed of sound An important property of fluids is the speed at which information (in particular, pressure changes due to the movement of an object) can travel within the fluid. This is equal to the average speed of molecules within the fluid, and it is called the speed of sound, noted c. It is important to quantify how fast the fluid is flowing relative to e speed of sound. For this, we define the Mach number [Ma] as the ratio of the local fluid speed V to the local speed of sound c: [Ma] V c (0/10) Since both V and c can be functions of space in a given flow, [Ma] may not be uniform (e.g. the Mach number around an airliner in flight is different at the nose than above the wings). Nevertheless, a single value is typically chosen to identify the Mach number of any given flow. It is observed that providing no heat or work transfer occurs, when fluids flow at [Ma] 0,3, their density ρ stays constant. Density variations in practice can be safely neglected below [Ma] = 0,6. These flows are termed incompressible. Above these Mach numbers, it is observed that when subjected to pressure variations, fluids exert work upon themselves, which translates into measurable density and temperature changes: these are called compressibility effects, and we will study them in chapter 9. In most flows, the density of liquids is almost invariant so that water flows are generally entirely incompressible. 15

16 When the fluid is air (and generally within a perfect gas), we will show in chapter 9 ( 9.3 p.196) that c depends only on the absolute temperature: c = γrt (0/11) in all cases for a perfect gas, where c is the local speed of sound (m s 1 ); γ is a gas property, approx. constant (dimensionless); and T is the local temperature (K) Viscosity We have accepted that a fluid element can deform continuously under pressure and shear efforts; however this deformation may not always be for free, i.e. it may require force and energy inputs which are not reversible. Resistance to strain in a fluid is measured with a property named viscosity. In informal terms, viscosity is the stickiness of fluids, i.e. their resistance to being sheared: honey and sugar syrups are very viscous while water has low viscosity. Formally, viscosity is defined as the ratio between the shear effort applied to a fluid particle, and the time rate of its deformation (its strain rate). For example, if a fluid element is strained uniformly in the x-direction, the velocity u x will be different everywhere inside the fluid element, and shear τ yx (the shear in the x-direction in the plane perpendicular to the y-direction) will occur. The viscosity µ is then defined as: where µ is the viscosity (N s m 2 or Pa s); τ yx is the shear in the x-direction along the plane perpendicular to the y-direction (Pa); and u x is the velocity in the x-direction (m s 1 ). µ = τ yx ( ux ) (0/12) y This equation 0/12 can be generalized for all directions (i, j and k) as: µ τ ji ( ui ) (0/13) j For most fluids, viscosity is the same in all directions and for all strain rates. This characteristic makes them Newtonian fluids. Viscosity is affected by temperature. We will study the effects of shear and viscosity in chapter 2. 16

17 0.5 Forces on fluids Fluids are subjected to, and subject their surroundings and themselves to forces. Identifying and quantifying those forces allows us to determine how they will flow. Three types of forces are relevant in fluid dynamics: gravity, pressure and shear Gravity Gravity (here, the attraction effort exerted on fluids at a distance by the Earth) is expressed with a vector д pointing downwards. Within the Earth s atmosphere, the magnitude д of this vector varies extremely slightly with altitude, and it may be considered constant at д = 9,81 m s 2. The weight force exerted on an object of mass m is then simply quantified as F weight = m д (0/14) Pressure The concept of pressure can be approached with the following conceptual experiment: if a flat solid surface is placed in a fluid at zero relative velocity, the pressure p will be the ratio of the perpendicular force F to the surface area A: p F (0/15) A where p is the pressure (N m 2 or pascals, 1 Pa 1 N m 2 ); F is the component of force perpendicular to the surface (N); and A is the surface area (m 2 ). Although in si units pressure is measured in Pascals (1 Pa 1 N m 2 ), in practice it is often measured in bars (1 bar Pa). Ambient atmospheric pressure at normal altitude varies with the weather and is approximately 1 bar. The pressure distribution within a fluid is related to their velocity distribution according to relations that we will study later. We shall then be looking for a pressure field p (x,y,z,t), a function of space an time. The pressure p exerting on an element of fluid can be thought of as the time- and space-average of the perpendicular component of impact force of its molecules on its neighbors. It is strictly a macroscopic property, that is, it cannot be defined at a microscopic level (there is no such thing as the pressure of a molecule ). In subsonic flows, it is a scalar property, meaning that for a given particle, it is the same in all directions. Pressure effects are explored in further detail in chapter Shear In the same thought experiment as above, shear τ expresses the efforts of a force parallel to a surface of interest: τ F A (0/16) where τ is the shear (N m 2 or Pa); F is the component of force parallel to the surface (N); 17

18 and A is the surface area (m 2 ). Contrary to pressure, shear is not a scalar, i.e. it can (and often does) take different values in different directions: on the flat plate above, it would have two components and could be represented by a vector τ = ( τ x,τ y ). We will explore shear in further detail in chapter Basic flow quantities A few fluid-flow related quantities can be quantified easily and are worth listing here. Mass flow is noted ṁ and represents the amount of mass flowing through a chosen surface per unit time. When the velocity across the surface is uniform, it can be quantified as: ṁ = ρv A (0/17) where ṁ is the mass flow (kg s 1 ); ρ is the fluid density (kg s 1 ); A is the area of the considered surface (m 2 ); and V is the component of velocity perpendicular to the surface (m s 1 ). Alternatively, one might consider instead the full fluid velocity and the area of a surface perpendicular to that velocity: ṁ = ρva (0/18) where V is the flow speed (m s 1 ); and A is the area of a surface perpendicular to the flow velocity (m 2 ). Volume flow is noted V and represents the volume of the fluid flowing through a chosen surface per unit time. Much like mass flow, when the velocity is uniform, it is quantified as: V = V A = VA = ṁ ρ where V is the volume flow (m 3 s 1 ). (0/19) Mechanical power is a time rate of energy transfer. Compression and expansion of fluids involves significant power. When the flow is incompressible (see 0.8 below), the mechanical power W necessary to force a fluid through a fixed volume where pressure losses occur is: W = F net, pressure V fluid, average = V p loss (0/20) where W is the power spent as work (W); and p is the pressure loss occurring due to fluid flow (Pa). Power as heat is also a time rate of energy transfer. When fluid flows uniformly and steadily through a fixed volume, its temperature T increases according to the net power as heat Q: Q = ṁ c fluid T (0/21) 18 where Q is the power spent as heat (W); T is the temperature change occurring in the fluid (K); and c fluid is the specific heat capacity of the fluid (J K 1 kg 1 ).

19 The heat capacity of fluids varies strongly according to the amount of work that they are performing. When no work is performed in a steady flow, the heat capacity is termed c p. In fluids such as liquid water and air, this capacity is almost independent from temperature. 0.7 Five equations The analytical and numerical resolution of problems in fluid mechanics is arched upon five important physical principles, which are sometimes called conservation principles. 1. Mass conservation: Without nuclear reactions, the total amount of matter at hand in a given phenomenon must remain constant. This statement can be expressed as: m system = cst dm system = 0 (0/22) dt 2. Conservation of linear momentum: This is a form of Newton s second law, which states that the net force F net Σ F applying to any given system of mass m is equal to the time change of its linear momentum m V : F net = d dt ( m V ) (0/23) 3. Conservation of angular momentum: This is a different form of Newton s second law, which states that the net moment M net, X Σ M X applied on a system about a point X is equal to the time change of its total angular momentum about this same point r m V : M net, X = d dt ( r m V ) (0/24) 4. Conservation of energy: This equation, also known as the first principle of thermodynamics, states that the total amount of energy within an isolated system must remain constant: de isolated system dt = 0 (0/25) 5. Second principle of thermodynamics: By contrast with all four other principles, this is not a conservation statement. Instead this principle states that the total amount of entropy S within a constant-energy system can only increase as time passes; in other words, the entropy change of any system must be greater than the heat transfer Q divided by the temperature T at which it occurs: ds system δq T (0/26) 19

20 These are the only five important equations written in fluid mechanics. We usually apply these statements to our problem in either one of two ways: We may wish to calculate the net (overall) change in fluid properties over a zone of interest so as to relate them to the net effect (forces, moments, energy transfers) of the fluid flow through the zone s boundaries. This method is called integral analysis and we will develop it in chapter 3. We may instead wish to describe the flow through our zone of interest in an extensive manner, aiming to obtain vector fields for the velocity and pressure everywhere, at all times. This method is called differential analysis and we will develop it in chapter Classification of fluid flows As we will see progressively, it is extremely difficult to obtain general solutions for fluid flow. Thus, whenever possible or reasonable, simplifying hypothesis are made about the behavior of any particular flow, that allow us to proceed with the analysis and obtain a reasonable, if inexact, specific solution. It is therefore a habit of fluid dynamicists to classify fluid flows in various categories, which are not necessarily incompatible. When approaching a given problem, we typically look out for the following characteristics: Time dependence Flows which do not vary with time are called steady. Steadiness is dependent on the chosen point of view: for example, the air flow around an airliner in cruise flight is seen to be steady from within the aircraft, but is obviously highly unsteady from the point of view of the air particles directly in the path of the airliner. Steadiness is also dependent on the selection of a suitable time frame: time-variations in a flow may be negligible if the time window used in the study is short enough. Mathematically, flows are steady when either the partial time derivative / t or total time derivative D/Dt (concepts we will study in chapter 4) of properties is zero: this considerably eases the search for solutions. It is important not to confuse steadiness (property fields constant in time) with uniformity (properties constant in space). To find out whether or not a flow is steady, compare instantaneous representations (photos, measurement readings, vector field representations, etc.) taken at different times. If they are all identical, the flow is steady. There does not exist an analytical method, however, that allows us to predict whether a flow will be steady. 20 Compressibility When the density of the fluid is uniform, the flow is said to be incompressible. The term is treacherous, because it refers to density, not pressure (incompressible flows almost always feature non-uniform pressure fields). To find out whether a gas flow is compressible, compute the Mach number (eq. 0/10 p.15). Below 0,3 the flow is always incompressible. Compressibility effects can be reasonably neglected below [Ma] = 0,6. Unless very specific phenomena such as phase changes or extreme speeds occur, the flow of liquids is always incompressible.

21 Temperature distribution In a fluid flow, temperature changes can occur due to three phenomena: Heat transfer from solid bodies; Changes in pressure and density due to work being performed on the fluid (by moving solid bodies, or by the fluid itself); Heat created through internal friction within the fluid. To find out whether a flow has uniform temperature (i.e. whether it is isothermal), three steps must be taken: 1. Quantify the heat transfer from external bodies. If that is zero, the flow is adiabatic, meaning there is no heat transfer. 2. Find out whether the flow is compressible. If not, the density changes are negligibly small and temperature changes do not occur due to compression or expansion. 3. Quantify the mechanical power lost by the fluid as it flows through the domain of interest. This power is integrally lost to heat. Because the heat capacity of fluids is generally very high, temperature changes due to this third factor are usually negligibly small. This is assessed with an example in exercise 0.6 p.25. Turbulence One last characteristic that we systematically attempt to identify in fluid flows is turbulence (or its opposite, laminarity). While laminar flows are generally very smooth and steady, turbulent flows feature multiple, chaotic velocity field variations in time and space. We shall first approach the concept of turbulence in chapter 5, and study it more formally in chapter 6. In the meantime, we may assess whether a flow will become turbulent or not using a non-dimensional parameter named the Reynolds number, noted [Re]: [Re] ρv L µ (0/27) To find out whether a flow will become turbulent, quantify [Re] by using the fluid properties ρ and µ, a representative fluid speed V, and a length L which is representative of the flow domain (for example, the stream-wise length of an obstacle). If the result is on the order of 10 4 or more, the flow is very likely to become turbulent over the length L. By contrast, with [Re] of the order of 10 2 or less, the flow is very likely to remain laminar over this length. This crude quantification has more meaning than it first appears. In particular, it assesses the influence of viscosity, which acts to to attenuate non-uniformities and unsteadiness in fluid flow, effectively damping its kinematics. We will explore the meaning behind this procedure, and apply it to concrete examples, in chapters 5 and following. 0.9 Limits of fluid mechanics Fluid mechanics is a complex discipline. It is easy to observe that flows as ordinary as sea waves crashing on a reef, water flowing down a river stream, or air blown into one s hands, display 21

22 tremendous geometrical complexity. Even after choosing to describe only the movement of macroscopic fluid particles instead of individual molecules (and thereby avoiding studying thousands of billions of individual movements), we still need to describe a three-dimensional field of properties (pressure, temperature, etc.), one of which, velocity, is itself three-dimensional. Thus, even before we begin describing the exact problem and a procedure to obtain its solution, we know that the mere description of a solution can have tremendous complexity. Additionally, we have to admit that in practice much progress can be made in the field of fluid mechanics. For example, our weather forecasts have almost no value beyond one week, and aircraft manufacturers with budgets measured in billions of dollars still make extensive use of wind tunnel models this despite our staggering continuous rate of progress in computing technology, and many decades of efforts dedicated to analytical fluid mechanics. In our present study of fluid mechanics, therefore, we shall proceed modestly, and will always take care to show the limits of our analysis. 22

23 Exercise sheet 0 (Important concepts) last edited April 1, 2018 Except otherwise indicated, we assume that fluids are Newtonian, and that: ρ water = kg m 3 ; p atm. = 1 bar; ρ atm. = 1,225 kg m 3 ; T atm. = 11,3 C; µ atm. = 1, N s m 2 ; д = 9,81 m s 2. Air is modeled as a perfect gas (R air = 287 J K 1 kg 1 ; γ air = 1,4; c pair = J kg 1 K 1 ). 0.1 Compressibility effects An aircraft is flying in air with density 0,9 kg m 3 and temperature 5 C. Above which flight speed would you expect the air flow over the wings to become compressible? 0.2 Pressure-induced force A 2 m by 2 m flat panel is used as the wall of a swimming pool (fig. 0.3). On the left side, the pressure is uniform at 1 bar. Figure 0.3 Pressure distribution on a flat plate Figure CC-0 o.c. 1. What is the pressure force exerted on the left side of the plate? On the right side of the plate, the water exerts a pressure which is not uniform: it increases with depth. The relation, expressed in pascals, is: p water = 1, , z (0/28) 2. What is the pressure force exerted on the right side of the plate? [Hint: we will explore the required expression in chapter 1 as eq. 1/14 p.35] 23

24 0.3 Shear-induced force A fluid flows over a 3 m by 3 m flat horizontal plate, in the x-direction as shown in fig Because of this flow, the plate is subjected to uniform shear τ zx = 1,65 Pa. Figure 0.4 Shear force exerting on a plate 1. What is the shear force applying on the plate? Figure CC-0 o.c. 2. What would be the shear force if the shear was not uniform, but instead was a function of x expressed (in pascals) as τ zx = 1,65 0,01 x 2? [Hint: we will explore the required expression in chapter 2 as eq. 2/20 p.53] 0.4 Speed of sound White [13] P1.87 Isaac Newton measured the speed of sound by timing the interval between observing smoke produced by a cannon blast and the hearing of the detonation. The cannon is shot 8,4 km away from Newton. What is the air temperature if the measured interval is 24,2 s? What is the temperature if the interval is 25,1 s? 0.5 Power lost to drag A truck moves with constant speed V on a road, with V = 100 km h 1. Because it experiences cross-wind, it is subjected to a drag F D with F D = 5 kn at an angle θ = 20, as shown in fig What is the power required to overcome drag? The drag force F D is applying at a distance 0,8 m behind the center of gravity of the truck. 2. What are the magnitude and the direction of the moment exerted by the drag F D about the center of gravity? Figure 0.5 Top view of a truck traveling at velocity V and subject to a drag force F D Figure CC-0 o.c. 24

25 0.6 Go-faster exhaust pipe The engine exhaust gases of a student s hot-rod car are flowing quasi-steadily in a cylindrical outlet pipe, whose outlet is slanted at an angle θ = 25 to improve the good looks of the car and provide the opportunity for an exercise. Figure 0.6 Exhaust gas pipe of a car. The outlet cross-section is at an angle θ relative to the axis of the pipe. Figure CC-0 o.c. Photo cropped, mirrored and edited from an original CC-by-sa by kazandrew2 The outlet velocity is measured at 15 m s 1, and the exhaust gas density is 1,1 kg m 3. The slanted outlet section area A is 420 cm What is the mass flow ṁ through the pipe? 2. What is the volume flow V of exhaust gases? Because of the shear within the exhaust gases, the flow through the pipe induces a pressure loss of 21 Pa (we will learn to quantify this in chapter 5). In these conditions, the specific heat capacity of the exhaust gases is c pgases = J kg 1 K What is the power required to carry the exhaust gases through the pipe? 4. What is the gas temperature increase due to the shear in the flow? 0.7 Acceleration of a particle Inside a complex, turbulent water flow, we are studying the trajectory of a cubic fluid particle of width 0,1 mm. The particle is accelerating at a rate of 2,5 m s 2. 25

26 1. What is the net force applying to the particle? 2. In practice, which types of forces could cause it to accelerate? 0.8 Flow classifications 1. Can an incompressible flow also be unsteady? 2. Can a very viscous fluid flow in a turbulent manner? 3. [more difficult] Can a compressible flow also be isothermal? 4. Give an example of an isothermal flow, of an unsteady flow, of a compressible flow, and of an incompressible flow. 26

27 Answers 0.1 If you adopt [Ma] = 0,6 as an upper limit, you will obtain V max = 709 km h 1 (eqs. 0/10 & 0/11 p.16). Note that propellers, fan blades etc. will meet compressiblity effects far sooner ) F left = 400 kn (eq. 0/15 p.17); 2) F right = 480 kn (eq. 1/14 p.35) ) F 1 = 14,85 N (eq. 0/16 p.17); 2) F 2 = 14,58 N (eq. 2/20 p.53) ,7 C & 5,6 C ) W = F drag V truck = 130,5 kw; 2) M = r F drag = N m, M = (points vertically upwards) ) ṁ = 0,2929 kg s 1 (eq. 0/17 p.18); 2) V = 266,2 L s 1 (eq. 0/19 p.18); 3) W = 5,59 W (eq. 0/20 p.18); 4) T = +0,0174 K (eq. 0/21 p.18), an illustration of remarks made in 0.8 p.21 regarding temperature distribution ) F net = 2, N (eq 0/23 p.19), such are the orders of magnitude involved in cfd calculations! 2) Only three kinds: forces due pressure, shear, and gravity ) yes, 2) yes if [Re] is high enough, 3) yes (in very specific cases such as high pressure changes combined with high heat transfer or high irreversibility, therefore generally no), 4) open the cap of a water bottle and turn it upside down: you have an isothermal, unsteady, incompressible flow. An example of compressible flow could be the expansion in a jet engine nozzle. 27

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29 Fluid Mechanics Chapter 1 Effects of pressure last edited September 10, Motivation Concept of pressure The direction of pressure Pressure on an infinitesimal volume Pressure in static fluids Fluid statics Shear in static fluids Pressure and depth Pressure distribution Hydrostatic pressure Atmospheric pressure Wall pressure forces and buoyancy Pressure forces on plane surfaces Buoyancy and Archimedes principle 37 Buoyancy and Archimedes principle Exercises 39 These lecture notes are based on textbooks by White [13], Çengel & al.[16], and Munson & al.[18]. 1.1 Motivation In fluid mechanics, only three types of forces apply to fluid particles: forces due to gravity, pressure, and shear. This chapter focuses on pressure (we will address shear in chapter 2), and should allow us to answer two questions: How is the effect of pressure described and quantified? What are the pressure forces generated on walls by static fluids? Video: pre-lecture briefing for this chapter by o.c. (CC-by) Concept of pressure We approached the concept of pressure in the introductory chapter with the notion that it represented force perpendicular to a given flat surface (eq. 0/15), for example a flat plate of area A: p F A (1/1) To appreciate the concept of pressure in fluid mechanics, we need to go beyond this equation The direction of pressure An important concept is that in continuum mechanics, the flat surface is imaginary. More precisely, a fluid is able to exert pressure on not only on solid surfaces, but also upon and within itself. In this context, we need to 29

30 rework eq. 1/1 so that now pressure is defined as perpendicular force per area on an infinitesimally small surface of fluid: p lim A 0 F A (1/2) Equation 1/2 may appear unsettling at first sight, because as A tends to zero, F also tends to zero; nevertheless, in any continuous medium, the ratio of these two terms tends to a single non-zero value. This brings us to the second particularity of pressure in fluids: the pressure on either side of the infinitesimal flat surface is the same regardless of its orientation. In other words, pressure has no direction: there is only one (scalar) value for pressure at any one point in space. Thus, in a fluid, pressure applies not merely on the solid surfaces of its container, but also everywhere within itself. We need to think of pressure as a scalar property field p (x,y,z,t) Pressure on an infinitesimal volume While pressure has no direction, it may not have the same value everywhere in a fluid, and so the gradient (the space rate of change) of pressure may not be null. For example, in a static water pool, pressure is uniform in the two horizontal directions, but it increases along with depth. Instead of a flat plate, let us now consider an infinitesimally small cube within the fluid (fig. 1.1). Because the cube is placed in a scalar field, the pressure exerting on each of its six faces may be different. The net effect of pressure will therefore have three components: one for each pair of opposing faces. Further down in this chapter, we will express this effect with the help of the gradient of pressure p, a vector field. For now, it is enough to notice that pressure itself is quantified as a (one-dimensional) scalar field, and the effect of pressure on a submerged volume as a (three-dimensional) vector field. 30 Figure 1.1 The pressure on each face of an infinitesimal volume may have a different value. The net effect of pressure will depend on how the pressure varies in space. These changes are labeled dp i in each of the i = x,y,z directions. figure CC-0 o.c.

31 1.3 Pressure in static fluids Fluid statics Fluid statics is the study of fluids at rest, i.e. whose velocity field V is everywhere null and constant: V V t = 0 = 0 (1/3) We choose to study this type of problem now, because it makes for a conceptually and mathematically simple case with which we can start analyzing fluid flow and calculate fluid-induced forces Shear in static fluids We will see in chapter 2 that shear efforts can be expressed as a function of viscosity and velocity: τ = µ u i x j i (2/16). We will also see that for most ordinary fluids (termed Newtonian fluids) the viscosity µ is independent from the velocity distribution. For now, it is enough to accept that in a static fluid (one in which u i x j = 0 since V (x,y,z,t) = 0), there cannot be any shear τ. Static fluids are unable to generate shear effort. This is a critical difference between solids and fluids. For example: A static horizontal beam extruding from a wall carries its own weight because it is able to support internal shear; but this is impossible for a static fluid. A solid brick resting on the ground exerts pressure on its bottom face, but because of internal shear, the pressure on its sides is zero. On the contrary, a static body of water of the same size and weight cannot exert internal shear. Because pressure has no direction, the body exerts pressure not only on the bottom face, but also laterally on the sides of its container Pressure and depth We now wish to express pressure within a fluid as a function of depth. Let us consider, inside a static fluid reservoir, a small element of fluid. We choose an element with finite horizontal surface S but infinitesimal height dz, as shown in fig The element s geometry is such that all of the lateral forces (which are due to pressure only) compensate each other. In the vertical direction, we are left with three forces, which cancel out because the element is not accelerating: F p bottom + F p top + m д = 0 p bottom S p top S mд = 0 31

32 Figure 1.2 A fluid element of (finite) horizontal surface S and infinitesimal height dz within a static fluid. figure CC-0 o.c. If we now rewrite the bottom pressure as p bottom = p top + dp, we obtain: (p top + dp)s p top S mд = 0 dp = m S д (1/4) This equation 1/4 expresses the fact that the infinitesimal variation of pressure dp between the top and bottom surfaces of the element is due only to its own weight. If we evaluate this variation along the infinitesimal height dz, we obtain: dp dz = dp dz m S dz д = m dv д = ρд (1/5) Thus, equation 1/5, which is obtained from Newton s second law applied to a static fluid, links the vertical variation in pressure to the fluid s density ρ and the local gravitational acceleration д. This equation is so useful that it is sometimes called the fundamental equation of fluid statics. If our coordinate system is not aligned with the vertical direction (and thus with gravity), we will have to adapt equation 1/5 and obtain a more general relationship: p x = ρд x p y = ρд y p z = ρд z (1/6) p x i + p y j + p k z = ρ д (1/7) We can simplify the writing of this last equation by using the mathematical operator gradient (see also Appendix A2 p.217), defined as so: i x + j y + k z (1/8) 32

33 And thus, equation 1/7 is more elegantly re-written as: p = ρ д (1/9) This equation is read the gradient of pressure is equal to the density times the gravitational acceleration vector. It expresses concisely a very important idea: in a static fluid, the spatial variation of pressure is due solely to the fluid s own weight. In chapter 4, we shall see that eq. 1/9 is a special case in a much more potent and complex relation called the Navier-Stokes equation (eq. 4/37 p.94). For now, it suffice for us to work with just pressure and gravity Pressure distribution An immediate consequence stemming from equation 1/9 is that in a static fluid (e.g. in a glass of water, in a swimming pool, in a calm atmosphere), pressure depends solely on height. Within a static fluid, at a certain altitude, we will measure the same pressure regardless of the surroundings (fig. 1.3). Figure 1.3 Pressure at a given depth (or height) in a static fluid does not depend on the environment. Here, as long as the fluid remains static, p A = p B = p C = p D. Figure CC-0 o.c Hydrostatic pressure How is pressure distributed within static liquid water bodies? The density of liquid water is approximately constant: ρ water = kg m 3. In a water reservoir, if we align the z-direction vertically downwards, equation 1/9 becomes: ( ) dp dz ( ) dp dz p = ρ д dp dz = ρд water water = ρ water д = ,81 = 9, Pa m 1 = 9, bar m 1 (1/10) 33

34 Therefore, in static water, pressure increases by approximately 0,1 bar/m as depth increases. For example, at a depth of 3 m, the pressure will be approximately 1,3 bar (which is the atmospheric pressure plus z dp dz ) Atmospheric pressure The density ρ air of atmospheric air is not uniform, since unlike for water it depends strongly on temperature and pressure. If we model atmospheric air as a perfect gas, once again orienting z vertically downwards, we can express the pressure gradient as: ( dp dz ) atm. = ρ air д = p 1 T д R (1/11) This time, the space variation of pressure depends on pressure itself (and it is proportional to it). A quick numerical investigation for ambient temperature and pressure (1 bar, 15 C) yields: ( dp dz ) atm. ambient = ,15 9, = 11,86 Pa m 1 = 1, bar m 1 This rate (approximately 0,1 mbar/m) is almost a thousand times smaller than that of water (fig. 1.4). Since the rate of pressure change depends on pressure, it also varies with altitude, and the calculation of pressure differences in the atmosphere is a little more complicated than for water. If we focus on a moderate height change, it may be reasonable to consider that temperature T, the gravitational acceleration д and the parameter R (gas constant) are uniform. In this admittedly restrictive case, equation 1/11 can Figure 1.4 Variation of pressure as a function of altitude for water and air at the surface of a water reservoir. The gradient of pressure with respect to altitude is almost a thousand times larger in water than in air. Figure CC-0 o.c. 34

35 be integrated as so: 2 1 dp dz = д p RT cst. 1 p dp = д 2 dz RT cst. 1 ln p 2 = д z p 1 RT cst. [ ] p 2 д z = exp p 1 RT cst. (1/12) Of course, air temperature varies significantly within the atmosphere (at moderate altitudes the change with altitude is approximately 6 K km 1 ). Adapting equation 1/12 for a uniform temperature gradient (instead of uniform temperature) is the subject of exercise 1.8 p.43. In practice, the atmosphere also sees large lateral pressure gradients (which are strongly related to the wind) and its internal fluid mechanics are complex and fascinating. Equation 1/12 is a useful and convenient model, but refinements must be made if precise results are to be obtained. 1.4 Wall pressure forces and buoyancy Pressure forces on plane surfaces When the pressure p exerted on a flat surface of area S is uniform, the resulting force F is easily easily calculated (F = p cst. S). In the more general case of a three-dimensional object immersed in a fluid with non-uniform pressure, the force must be expressed as a vector and obtained by integration: F pressure = d F = p n ds (1/13) S where the S-integral denotes an integration over the entire surface; and n is a unit vector describing, on each infinitesimal surface element ds, the direction normal to the surface. S Video: when pressure-induced forces in static fluids matter: 24 hours of heavy tonnage transit through the Miraflores locks in Panama. Can you quantify the force applying on a single lock door? by Y:XyliboxFrance (styl) Equation 1/13 is easily implemented in software algorithms to obtain numerically, for example, the pressure force resulting from fluid flow around a body such as an aircraft wing. In our academic study of fluid mechanics, however, we will restrict ourselves to the simpler cases where the surface is perfectly flat (and thus n is everywhere the same). Equation 1/13 then becomes: F pressure = df = p ds (1/14) for a flat surface. S What is required to calculate the scalar F in eq. 1/14 is an expression of p as a function of S. In a static fluid, this expression will be given by eq. 1/9. Typically in two dimensions x and y we re-write ds as ds = dx dy and we may then proceed with the calculation starting from F pressure = p (x,y) dx dy (1/15) S 35

36 Such a calculation gives us the magnitude of the pressure force, but not its position. This position can be evaluated by calculating the magnitude of moment generated by the pressure forces about any chosen point X. This moment M X, using notation shown in fig. 1.5, is expressed as: M X = d M X = r XF d F = r XF p n ds (1/16) S S where r XF is a vector expressing the position of each infinitesimal surface relative to point X. Much like eq. 1/13 above, this eq. 1/16 is easily implemented in software algorithms but not very approachable on paper. In the simple case where we study only a flat surface, and where the reference point X is in the same plane as the surface, eq. 1/16 simplifies greatly and we can calculate the magnitude M X as: M X = dm X = r XF df = r XF p ds (1/17) S for a flat surface, with X in the plane of the surface. S S S Once both F pressure and M X pressure have been quantified, the distance R XF between point X and the application point of the net pressure force is easily computed: R XF = M X pressure F pressure (1/18) While we restrict ourselves to simple cases where surfaces are flat, the process above outlines the method used by computational fluid dynamics (cfd) software packages to determine the position of forces resulting from fluid pressure on any arbitrary solid object. 36 Figure 1.5 Moment generated about an arbitrary point X by the pressure exerted on an arbitrary surface (left: perspective view; right: side view). The vector n is a convention unit vector everywhere perpendicular to the infinitesimal surface ds considered. Figure CC-0 o.c.

37 1.4.2 Buoyancy and Archimedes principle Any solid body immersed within a fluid is subjected to pressure on its walls. When the pressure is not uniform (for example because the fluid is subjected to gravity, although this may not be the only cause), then the net force due to fluid pressure on the body walls will be non-zero. When the fluid is purely static, this net pressure force is called buoyancy. Since in this case, the only cause for the pressure gradient is gravity, the net pressure force is oriented upwards. The buoyancy force is completely independent from (and may or may not compensate) the object s weight. Since it comes from equation 1/9 that the variation of pressure within a fluid is caused solely by the fluid s weight, we can see that the force exerted on an immersed body is equal to the weight of the fluid it replaces (that is to say, the weight of the fluid that would occupy its own volume were it not there). This relationship is sometimes named Archimedes principle. The force which results from the static pressure gradient applies to all immersed bodies: a submarine in an ocean, an object in a pressurized container, and of course, the reader of this document as presently immersed in the earth s atmosphere. Video: playing around with an air pump and a vacuum chamber by Y:Roobert33 (styl) Figure 1.6 Immersion in a static fluid results in forces that depend on the body s volume. They can evidenced by the removal of the fluid (for example in a depressurized semi-spherical vessel). Figure CC-0 o.c. 37

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39 Exercise sheet 1 (Effects of pressure) last edited April 18, 2018 These lecture notes are based on textbooks by White [13], Çengel & al.[16], and Munson & al.[18]. Except otherwise indicated, we assume that fluids are Newtonian, and that: ρ water = kg m 3 ; p atm. = 1 bar; ρ atm. = 1,225 kg m 3 ; T atm. = 11,3 C; µ atm. = 1, N s m 2 ; д = 9,81 m s 2. Air is modeled as a perfect gas (R air = 287 J K 1 kg 1 ; γ air = 1,4; c pair = J kg 1 K 1 ). 1.1 Pressure in a fluid A small water container whose geometry is described in fig. 1.7 is filled with water. What is the pressure at the bottom of the container? Figure 1.7 A small water container. Figure CC-0 o.c. 1.2 Pressure measurement A tube is connected to a pressurized vessel as shown in fig The U-tube is filled with water. What is the pressure p int. in the vessel? Figure 1.8 Working principle of a simple liquid tube manometer. The outlet is at atmospheric pressure p atm.. Figure CC-0 o.c. What would be the height difference shown for the same internal pressure if mercury (ρ mercury = kg m 3 ) was used instead of water? 39

40 1.3 Straight water tank door A water tank has a window on one of its straight walls, as shown in figure 1.9. The window is 3 m high, 4 m wide, is positioned 0,4 m above the bottom of the tank. Figure 1.9 An aquarium tank with a window installed on one of its walls Figure CC-0 o.c. 1. What is the force applying on the window due to the pressure exerted by the water? 2. At which height does this force apply? 1.4 Water canal access door A water canal used in a laboratory is filled with stationary water (fig. 1.10). An observation window is installed on one of the walls of the canal, to enable observation and measurements. The window is hinged on its bottom face. The hinge stands 1,5 m below the water surface. The window has a length of 0,9 m and a width of 2 m. The walls of the canal are inclined with an angle θ = 60 relative to horizontal. Figure 1.10 A door installed on the wall of a water canal. The water in the canal is perfectly still. Figure CC-0 o.c. 1. Represent graphically the pressure of the water and atmosphere on each side of the window What is the moment exerted by the pressure of the water about the axis of the window hinge?

41 3. If the same door was positioned at the same depth, but the angle θ was decreased, would the moment be modified? (briefly justify your answer, e.g. in 30 words or less) 1.5 Water lock A system of lock doors is set up to allow boats to travel up the side of a hill (figure 1.11). Figure 1.11 Outline schematic of a water canal lock. Figure 1 CC-0 o.c.; figure 2 CC-by-sa o.c. We are studying the force and moment exerted by the water on the lock door labeled A. At this instant, the water levels are as shown in figure 1.11: 2 m in the lower canal, 5 m in the lock, and 9 m in the upper canal. The door is 2,5 metres wide. 41

42 1. Sketch the distribution of the pressure exerted by the water and the atmosphere on both sides of door A. 2. What is the net force exerted by the water on door A? 3. At which height is this force exerting? 1.6 Buoyancy force on a tin can A student contemplates a tin can of height 10 cm and diameter 7 cm. 1. If one considers that the atmospheric density is uniform, what is the buoyancy force generated by the atmosphere on the can when it is positioned vertically? 2. What is the force generated when the can is immersed in water at a depth of 20 cm? At a depth of 10 m? 3. What is the buoyancy force generated when the can is immersed in the water in a horizontal position? Instead of uniform density, we now wish to calculate the atmospheric buoyancy under the hypothesis of uniform temperature (room temperature 20 C). 4. Starting from equation 1/9: p = ρ д, show that when the temperature Tcst. is assumed to be uniform, the atmospheric pressure p at two points 1 and 2 separated by a height difference z is such that: p 2 p 1 = exp [ д z RT cst. 5. What is the buoyancy generated by the room atmosphere? ] (1/12) 1.7 Buoyancy of a barge A barge of very simple geometry is moored in a water reservoir (fig. 1.12). Figure 1.12 Basic layout of a barge floating in water. Figure CC-0 o.c. 1. Sketch the distribution of pressure on each of the immersed walls of the barge (left and right sides, rear, bottom and slanted front). 2. What is the force resulting from pressure efforts on each of these walls? What is the weight of the barge?

43 1.8 Atmospheric pressure distribution The integration we carried out in p.34 to model the pressure distribution in the atmosphere was based on the hypothesis that the temperature was uniform and constant (T = T cst. ). In practice, this may not always be the case. 1. If the atmospheric temperature decreases with altitude at a constant rate (e.g. of 7 K km 1 ), how can the pressure distribution be expressed analytically? A successful fluid dynamics lecturer purchases an apartment at the top of the Burj Khalifa tower (800 m above the ground). Inside the tower, the temperature is controlled everywhere at 18,5 C. Outside, the ground temperature is 30 C and it decreases linearly with altitude (gradient: 7 K km 1 ). A door is opened at the bottom of the tower, so that at zero altitude the air pressure (1 bar) is identical inside and outside of the tower. For the purpose of the exercise, we pretend the tower is entirely hermetic (meaning air is prevented from flowing in or out of its windows). 2. What is the pressure difference between each side of the windows in the apartment at the top of the tower? 1.9 Reservoir door Munson & al. [18] 2.87 A water reservoir has a door of width 3 m which is held in place with a horizontal cable, as shown in fig The door has a mass of 200 kg and the friction in the hinge is negligible. Figure 1.13 A sealed, hinged door in a water reservoir. The width across the drawing (towards the reader) is 3 m Figure CC-0 o.c. 1. What is the force in the cable? 2. If the water height was decreased, how would this force be modified? (briefly justify your answer, e.g. in 30 words or less) 43

44 1.10 Atmospheric buoyancy force non-examinable. CC-0 o.c. If the atmospheric density is considered uniform, estimate the buoyancy force exerted on an Airbus A380, both on the ground and in cruise flight (ρ cruise = 0,4 kg m 3,T cruise = 40 C). Figure 1.14 Airbus A Drawing CC-by-sa by Julien Scavini Length overall 72,73 m Wingspan 79,75 m Height 24,45 m Wing area 845 m 2 Aspect ratio 7,5 Wing sweep 33,5 Maximum take-off weight kg Typical operating empty weight kg Table 1.1 Characteristics of the Airbus A Lock with diagonally-mounted doors non-examinable. CC-0 o.c. The doors of the lock studied in exercise 1.5 p.41 are replaced with diagonally-mounted doors at an angle α = 20, mounted such that no bending moment is sustained by the hinges (fig. 1.15). What is the force exerted by door A on the other door? Figure 1.15 Lock doors mounted with an angle relative one to another. The angle relative to the case in exercise 1.5 is α = 20. The width of the canal is still 5 m. Figure CC-0 o.c. 44

45 Answers 1.1 p A = p atm. + 0,039 bar 1,039 bar ) p inside = p atm. + 0,0157 bar 1,0157 bar; 2) z 2 = 1,1765 cm ) F net = ρдl ( Z max L max 1 2 L2 max) = 176,6 kn; 2) M net, bottom hinge = ρдl ( 1 2 Z maxl 2 max 1 3 max) L3 = 176,6 kn m so the force exerts at R = M net F net = 1 m above the bottom hinge ) M net = 7,79 kn m; 3) observe the equation used to calculate M net to answer this question. If needed, ask for help in class! 1.5 2) F net, door A = ρдl [ 1 2 z2] 9 m = +686,7 kn (positive in right direction) 5 m L 3) M net, door A, about bottom axis = F net 2 = 2,4689 MN m (positive clockwise): F net, door A exerts at R 2 = 3,595 m from the bottom ) F vertical = 4,625 mn upwards; 2) F vertical = 3,775 N in both cases; 3) There is no change; 4) See 1.3.6; 5) F vertical = 4,487 mn upwards (and so in question 1 we were off by 3 %) ) F rear = 0,2453 MN, F side = 2,1714 MN, F bottom = 13,734 MN, F front = 0,3468 MN; 3) F buoyancy = 13,979 MN (1 425 t) ) p 2 p 1 = ( 1 + kz M water = L ( dp dz T 1 ) д ) kr. water [ H 2 r 2 sin θ 3 r 3] R 2 = 0,4186 MN m; so, F cable = 81,13 kn Assuming a volume of approx m 3, we obtain approx. 30,9 kn on the ground, 10,4 kn during cruise The moment is brought to zero by an inter-door force F sideways = 1,068 MN (perpendicular to the canal axis). 45

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47 Fluid Mechanics Chapter 2 Effects of shear last edited April 18, Motivation Concept of shear The direction of shear Shear on an infinitesimal volume Slip and viscosity The no-slip condition Viscosity Newtonian Fluid Wall shear forces Exercises 55 These lecture notes are based on textbooks by White [13], Çengel & al.[16], and Munson & al.[18]. 2.1 Motivation In fluid mechanics, only three types of forces apply to fluid particles: forces due to gravity, pressure, and shear. This chapter focuses on shear, and should allow us to answer two questions: How is the effect of shear described and quantified? What are the shear forces generated on walls by simple flows? Video: pre-lecture briefing for this chapter by o.c. (CC-by) Concept of shear We approached the concept of shear in the introductory chapter with the notion that it represented force parallel to a given flat surface (eq. 0/16), for example a flat plate of area A: τ F A (2/1) Like we did with pressure, to appreciate the concept of shear in fluid mechanics, we need to go beyond this equation The direction of shear Already from the definition in eq. 2/1 we can appreciate that parallel to a flat plate can mean a multitude of different directions, and so that we need more than one dimension to represent shear. Furthermore, much in the same way as we did for pressure, we do away with the flat plate and accept that shear is a field, i.e. it is an effort applying not only upon solid objects but also upon and within fluids themselves. We replace eq. 2/1 with a more general definition: F τ lim A 0 A (2/2) 47

48 Contrary to pressure, shear is not a scalar, i.e. it can (and often does) take different values in different directions. At a given point in space we represent it as a vector τ = ( τ x,τ y,τ y ), and in a fluid, there is a shear vector field: Video: cloud movements in a time-lapse video on an interesting day are evidence of a highlystrained atmosphere: pilots and meteorologists refer to this as wind shear. by Y:StormsFishingNMore (styl) τ (x,y,z,t) τ x τ y (2/3) τ z (x,y,z,t) Shear on an infinitesimal volume Describing the changes in space of the shear vector field requires another mathematical dimension (called order). Instead of a flat plate, let us consider an infinitesimally small cube within the fluid (fig. 2.1). Because the cube is immersed inside a vector field, the shear vector exerting on each of its six faces may be different. In order to express the efforts on any given face, we express a component of shear with two subscripts, the first indicating the direction normal to the surface of interest, and the second indicating the direction of the effort. For example, τ xy represents the shear in the y-direction on a surface perpendicular to the x-direction. On this face, the shear vector would be: τ xj = τ xx + τ xy + τ xz (2/4) = τ xx i + τ xy j + τ xz k (2/5) where the subscript xj indicates all of the directions (j = x,y,z) on a face perpendicular to the x-direction. In eq. 2/4, the reader may be surprised to see the term τ xx appear a shear effort perpendicular to the surface of interest. This is because the faces of the infinitesimal cube studied here (shown in fig. 2.1) are not solid. They are permeable, and the local velocity on each one may (in fact, must, if there is to be any flow) include a component of velocity through the face of the cube. Thus, there is no reason for the shear effort, which is three-dimensional, to be aligned along each flat surface. As the fluid travels across any face, it can be sheared (which results in strain) in any arbitrary direction, regardless of the local pressure and thus shear can and most often does have a component (τ ii ) perpendicular to an arbitrary surface inside a fluid. Figure 2.1 Shear efforts on a cubic fluid particle (with only the efforts on the visible faces 1 to 3 represented). The shear tensor τ ij has six members of three components each. Figure CC-0 o.c. 48

49 Now, the net shear effect on the cube will have eighteen components: one tree-dimensional vector for each of the six faces. Each of those components may take a different value. The net shear could perhaps be represented as en entity a tensor containing six vectors τ 1, τ 2, τ 3... τ 6. By convention, however, shear is notated using only three vector components: one for each pair of faces. Shear efforts on a volume are thus represented with a tensor field τ ij : τ ij τ xj τ yj τ zj τ ij τ xj {1,4} τ yj {2,5} τ zj {3,6} τ xx τ xy τ xz τ yx τ yy τ yz τ zx τ zy τ zz (2/6) In this last equation 2/6, each of the nine components of the tensor acts as the container for two contributions: one for each of the two faces perpendicular to the direction expressed in its first subscript. So much for the shear effort on an element of fluid. What about the net force due to shear on the fluid element? Not every element counts: part of the shear will accelerate (change the velocity vector) the particle, while part of it will merely strain (deform) the particle. Quantifying this force thus requires making a careful selection within the eighteen components of τ ij. We may start with the x-direction, which consists of the sum of the component of shear in the x-direction on each of the six cube faces: F shear x = S 3 τ zx 3 S 6 τ zx 6 +S 2 τ yx 2 S 5 τ yx 5 +S 1 τ xx 1 S 4 τ xx 4 (2/7) Given that S 3 = S 6 = dx dy, that S 2 = S 5 = dx dz and that S 1 = S 4 = dz dy, this is re-written as: F shear x = dx dy( τ zx 3 τ zx 6 ) + dx dz( τ yx 2 τ yx 5 ) + dz dy( τ xx 1 τ xx 4 ) (2/8) In the same way we did with pressure in chapter 1 ( p.31), we express each pair of values as derivative with respect to space multiplied by an infinitesimal distance: F shear x = dx dy = dv ( dz τ ) ( zx + dx dz z ( τzx z + τ yx y + τ ) xx x dy τ yx y ) + dz dy ( dx τ ) xx x (2/9) 49

50 If we make use of the operator divergent (see also Appendix A2 p.217), written : A A x x A ij x i + y j + z k (2/10) A xx x A xy x A xz x + A y y + A yx y + A yy y + A yz y + A z z + A zx z + A zy z + A zz z = A ix A iy A iz (2/11) (2/12) we can re-write eq. 2/9 and see that the net shear force in the x-direction is equal to the particle volume times the divergent of the shear in the x- direction: F shear x = dv τ ix (2/13) The y- and z-direction are taken care of in the same fashion, so that we can gather up our puzzle pieces and express the force per volume due to shear as the divergent of the shear tensor: F shear = F shear x F shear y = dv F shear z τ ix τ iy τ iz = dv τ ij (2/14) This exploration goes beyond the theory required to to through this chapter (we will come back to it when we start concerning ourselves with the dynamics of fluid particles in chapter 4, where the divergent of shear will make part of the glorious Cauchy equation). It suffices for now to sum up our findings as follows: Shear at a point in space has three components it is a vector field; The effect of shear on a volume of fluid has eighteen components it is a second-order tensor field; The net force due to shear on a volume of fluid, expressed using the divergent of the shear tensor, has three components it is a vector field. 2.3 Slip and viscosity The no-slip condition We observe that whenever we measure the velocity of a fluid flow along a solid wall, the speed tends to zero as we approach the wall surface. In other words, the fluid sticks to the surface regardless of the overall faraway flow velocity. This phenomenon is called the no-slip condition and is of paramount importance in fluid mechanics. One consequence of this is that fluid flows near walls are dominated by viscous effects (internal friction) due to the large velocity gradients. 50

51 2.3.2 Viscosity When a solid wall is moved longitudinally within a fluid, the fluid generates an opposing friction force through viscous effects (fig. 2.2). We call viscosity µ the ratio between the fluid velocity gradient and the shear effort. For example, the norm of the shear τ xy on a surface perpendicular to the x-direction, in the y-direction, can be expressed as: τ xy = µ u y x = µ v x In the general case, viscosity is defined as the (scalar) ratio between the norm of shear and the corresponding strain rate. The strain rate corresponding to the shear in the j-direction is the rate of change in the i-direction of the velocity in the j-direction ( u j / i): µ τ ij ) (2/15) ( uj i τ ij = µ u j i (2/16) in which the subscript i is an arbitrary direction (x, y or z) and j is the direction following it in order (e.g. j = z when i = y); and where µ is the viscosity (or dynamic viscosity) (Pa s). Viscosity µ is measured in Pa s, which is the same as N s m 2 or kg m 1 s 1. It has historically been measured in poise (1 poise 0,1 Pa s). Figure 2.2 Any velocity gradient u y x in the flow results in a shear force F in the direction y. The ratio between the shear and the velocity gradient is called viscosity. Figure CC-0 o.c. 51

52 The concept of kinematic viscosity ν (Greek letter nu) is sometimes used instead of the dynamic viscosity; it is defined as ν µ ρ (2/17) where ν is measured in m 2 s Newtonian Fluid Fluids for which µ is independent from u j i are called Newtonian fluids. Most fluids of interest in engineering fluid mechanics (air, water, exhaust gases, pure gases) can be safely modeled as Newtonian fluids. Their viscosity µ varies slightly with pressure and mildly with temperature in our study of fluid mechanics, we will not take these dependencies into account. The values of viscosity vary very strongly from one fluid to another: for example, honey is roughly ten thousand times more viscous than water, which is roughly a hundred times more viscous than ambient air. The viscosities of various fluids are quantified in fig p.119. Oil-based paint, blood and jelly-based fluids are strongly non-newtonian; they require more complex viscosity models (fig. 2.3). 52 Figure 2.3 Various possible viscosity characteristics of fluids. Those for which µ is independent of u j / i are called Newtonian. Figure CC-0 o.c.

53 2.4 Wall shear forces The calculation of forces due to shear is very similar to that which we used in the previous chapter with pressure. When the shear τ exerted on a flat surface of area S is uniform, the resulting force F is easily easily calculated (F = τ cst. S). In the more general case of a three-dimensional object immersed in a fluid with non-uniform shear, the force must be expressed as a vector and obtained by integration: F shear = d F = τ surface ds (2/18) S Much like equation 1/13 in the previous chapter, eq. 2/18 is easily implemented in software algorithms to obtain numerically, for example, the force resulting from shear due to fluid flow around a body such as an aircraft wing. In our academic study of fluid mechanics, however, we will restrict ourselves to the simpler cases where the surface is perfectly flat and the shear has uniform direction. The force in eq. 2/18, exterting in the i-direction due to shear on a surface perpendicular to the j-direction, then becomes: F shear ji = df = τ ji ds (2/19) S What is required to calculate the scalar F in eq. 2/19 is an expression of τ as a function of S. We proceed like we did in the previous chapter, splitting ds as ds = di dk before proceeding with the calculation starting from; F shearji = S S τ ji (i,k) di dk (2/20) = µ ui j di dk (2/21) where i is the direction of the force; j is the direction perpendicular to the flat surface; and k is the third (orthonormal) direction. The above expression 2/21 is perhaps easier to read when it is developed. For example, the shear force F shearyi exerting on a plate perpendicular to the y-direction is: F shearyi = F shearyx 0 F shearyz with u F shearyx = µ dx dz y w F shearyz = µ dx dz y Just like for pressure, the moment M Xj generated about a point X by the shear forces in the i-direction on a plane surface perpendicular to the j-direction can be calculated as M Xj = d M X = r XF d F = r XF τ ij ds (2/22) S S S 53

54 If the point X is in the plane of the surface of interest, this can be expressed as: u i M Xj = µ r XF di dk (2/23) j u x M Xy = µ r XF dx dz (2/24) y 54

55 Exercise sheet 2 (Effects of shear) last edited April 30, 2017 These lecture notes are based on textbooks by White [13], Çengel & al.[16], and Munson & al.[18]. Except otherwise indicated, we assume that fluids are Newtonian, and that: ρ water = kg m 3 ; p atm. = 1 bar; ρ atm. = 1,225 kg m 3 ; T atm. = 11,3 C; µ atm. = 1, N s m 2 ; д = 9,81 m s 2. Air is modeled as a perfect gas (R air = 287 J K 1 kg 1 ; γ air = 1,4; c pair = J kg 1 K 1 ). 2.1 Flow in between two plates Munson & al. [18] Ex1.5 A fluid is forced to flow between two stationary plates (fig. 2.4). We observe that the flow is laminar (smooth and fully steady), with a univorm velocity profile u = f (y) which is linked to the average fluid velocity V average by the relationship: u = 3 2 V average [ ( y ) 2 ] 1 H (2/25) where y is measured from the middle of the gap; and H is half of the gap length. Figure 2.4 Velocity distribution for laminar flow in between two plates, also known as Couette flow. Figure CC-0 o.c. The fluid viscosity is 2 N s m 2, the average velocity is 0,6 m s 1 and the two plates are 10 mm apart. 1. What is the shear effort τ yx plate generated on the lower plate? 2. What is the shear effort τ yx in the middle plane of the flow? 55

56 2.2 Friction on a plate A plate the size of an A4 sheet of paper (210 mm 297 mm) is moved horizontally at constant speed above a large flat surface (fig. 2.5). We assume that the velocity profile of the fluid betweeen the plate and the flat surface is entirely uniform, smooth, and steady. Figure 2.5 A plate moved horizontally across a flat surface. Figure CC-0 o.c. 1. Express the force F yz due to shear on the plate as a function of its velocity U plate, the gap height H, and the properties of the fluid. 2. The plate speed is U plate = 1 m s 1 and the gap height is H = 5 mm. What is the shear force if the fluid is air (µ atm. = 1, N s m 2 ), and if the fluid is honey (µ honey = 40 N s m 2 )? 3. If a very long and thin plate with the same surface area was used instead of the A4-shaped plate, would the shear force be different? (briefly justify your answer, e.g. in 30 words or less) 2.3 Viscometer Çengel & al. [16] 2-78 An instrument designed to measure the viscosity of fluids (named viscometer) is made of two coaxial cylinders (fig. 2.6). The inner cylinder is immersed in a liquid, and it rotates within the stationary outer cylinder. The two cylinders are 75 cm tall. The inner cylinder diameter is 15 cm and the spacing is 1 mm. When the inner cylinder is rotated at 300 rpm, a friction-generated moment of 0,8 N m is measured. 1. If the flow in between the cylinders corresponds to the simplest possible flow case (steady, uniform, fully-laminar), what is the viscosity of the fluid? 2. Would a non-newtonian fluid induce a higher moment? (briefly justify your answer, e.g. in 30 words or less) 56 [Note: in practice, when the inner cylinder is turned at high speed, the flow displays mesmerizing patterns called Taylor Couette vortices, the description of which is much more complex!]

57 Figure 2.6 Sketch of a cylinder viscometer. The width of the gap has been greatly exaggerated for clarity. Figure CC-0 o.c. 2.4 Boundary layer White [13] P1.56 A laminar fluid flow occurs along a wall (fig. 2.7). Close to the wall (y < δ), we observe that viscous effects dominate the mechanics of the flow. This zone is designated boundary layer. The speed u (y) can then be modeled with the relation: u = U sin ( πy ) 2δ in which U is the flow speed far away from the wall. (2/26) Figure 2.7 Velocity profile across the boundary layer. Figure CC-0 o.c. The fluid is helium at 20 C and 1 bar ( N s m 2 ); measurements yield U = 10,8 m s 1 and δ = 3 cm. 1. What is the shear effort τ yx on the wall? 2. At which height y 1 above the surface will the shear effort be half of this value? 3. What would be the wall shear if helium was replaced with water ( kg m 1 s 1 )? 57

58 2.5 Clutch Çengel & al. [16] 2-74 Two aligned metal shafts are linked by a clutch, which is made of two disks very close one to another, rotating in the same direction at similar (but not identical) speeds. The disk diameters are both 30 cm and the gap between them is 2 mm ; they are submerged in Sae30w oil with viscosity 0,38 kg m 1 s 1. Figure 2.8 Sketch of the two disks constituting the clutch. The gap width has been exaggerated for clarity. Figure CC-0 o.c. The power shaft rotates at rpm, while the powered shaft rotates at rpm. We consider the simplest possible flow case (steady, laminar) in between the two disks. 1. What is the moment imparted by one disk to the other? 2. How would the moment change if the radius of each disk was doubled? 3. What is the transmitted power and the clutch efficiency? 4. Briefly (e.g. in 30 words or less) propose one reason why in practice the flow in between the two disks may be different from the simplest-case flow used in this exercise. 58

59 Answers 2.1 1) τ yx y= H = 720 N m 2 ; 2) τ yx y=0 = 0 N m ) F yz = L 1 L 2 µ U H = 1, N for air, and 500 N for honey(!). 2.3 µ = hm 2πωR 3 1 H = 1, N s m ) τ yx,wall = µ πu 2δ = 5, N 2 m 1 for helium; 2) y 1 = 2 3 δ = 2 cm; 3) τ yx,wall = 0,565 N m 2 for water ) M = π µω 2 h R4 = 0,8228 N m; 3) W W 2 = ω 2 M = 120 W; η clutch = 2 W = 1 96,4 % (remember this is a very low-power, low-relative speed, laminar-flow case). 59

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61 Fluid Mechanics Chapter 3 Analysis of existing flows last edited September 23, Motivation The Reynolds transport theorem Control volume Rate of change of an additive property 61 Rate of change of an additive property Mass conservation Change of linear momentum Change of angular momentum Energy conservation The Bernoulli equation Limits of integral analysis Exercises 73 These lecture notes are based on textbooks by White [13], Çengel & al.[16], and Munson & al.[18]. 3.1 Motivation Our objective for this chapter is to answer the question what is the net effect of a given fluid flow through a given volume?. Here, we develop a mass, momentum and energy accounting methodology to analyze the flow of continuous medium. This method is not powerful enough to allow us to describe extensively the nature of fluid flow around bodies; nevertheless, it is extremely useful to quantify forces, moments, and energy transfers associated with fluid flow. Video: pre-lecture briefing for this chapter, part 1/2 by o.c. (CC-by) The Reynolds transport theorem Control volume Let us begin by describing the flow which interests us a a generic velocity field V = (u,v,w) which is a function of space and time ( V = f (x,y,z,t)). Within this flow, we are interested in an arbitrary volume named control volume (CV) which is free to move and change shape (fig. 3.1). We are going to measure the properties of the fluid at the borders of this volume, which we call the control surface (CS), in order to compute the net effect of the flow through the volume. At a given time, the control volume contains a certain amount of mass which we call the system (sys). Thus the system is a fixed amount of mass transiting through the control volume at the time of our study, and its properties (volume, pressure, velocity etc.) may change in the process. All along the chapter, we are focusing on the question: based on measured fluid properties at some point in space and time (the properties at the control surface), how can we quantify what is happening to the system (the mass inside the control volume)? 61

62 Figure 3.1 A control volume within a flow. The system is the amount of mass included within the control volume at a given time. At a later time, it may have left the control volume, and its shape and properties may have changed. The control volume may also change shape with time, although this is not represented here. Figure CC-0 o.c Rate of change of an additive property In order to proceed with our calculations, we need a robust accounting methodology. We start with a dummy fluid property B, which we will later replace with physical variables of interest. Let us thus consider an arbitrary additive property B of the fluid. By the term additive property, we mean that the total amount of property is divided if the fluid is divided. For instance, this is true of mass, volume, energy, entropy, but not pressure or temperature. The specific (i.e. per unit mass) value of B is designated b B/m. We now wish to compute the variation of a system s property B based on measurements made at the borders of the control volume. We will achieve this with an equation containing three terms: The time variation of the quantity B within the system is measured with the term db sys dt. This may represent, for example, the rate of change of the fluid s internal energy as it travels through a jet engine. Within the control volume, the enclosed quantity B CV can vary by accumulation (for example, mass may be increasing in an air tank fed with compressed air): we measure this with the term db CV dt. Finally, a mass flux may be flowing through the boundaries of the control volume, carrying with it some amount of B every second: we name that net flow out of the system Ḃ net Ḃ out Ḃ in. 62

63 We can now link these three terms with the simple equation: db sys dt = db CV dt + Ḃ net (3/1) the rate of change of B for the system the rate of change = of B within the control volume + the net flow of B through the boundaries of the control volume Since B may not be uniformly distributed within the control volume, we like to express the term db CV dt as the integral of the volume density B V with respect to volume: db CV = d B dt dt CV V dv = d ρb dv (3/2) dt CV Obtaining a value for this integral may be difficult, especially if the volume of the control volume CV is itself a function of time. The term Ḃ net can be evaluated by quantifying, for each area element da of the control volume s surface, the surface flow rate ρbv of property B that flows through it (fig. 3.2). The integral over the entire control volume surface CS of this term is: Ḃ net = ρbv da = ρb ( V rel n) da (3/3) CS where flows and velocities are positive outwards and negative inwards by convention, CV is the control volume, CS is the the control surface (enclosing the control volume), n is a unit vector on each surface element da pointing outwards, V rel is the local velocity of fluid relative to the control surface, and V V rel n is the local cross-surface speed. By inserting equations 3/2 and 3/3 into equation 3/1, we obtain: CS Figure 3.2 Part of the system may be flowing through an arbitrary piece of the control surface with area da. The n vector defines the orientation of da surface, and by convention is always pointed outwards. Figure CC-0 o.c. 63

64 db sys dt = d dt ρb dv + CV CS ρb ( V rel n) da (3/4) Equation 3/4 is named the Reynolds transport theorem; it stands now as a general, abstract accounting tool, but as we soon replace B by meaningful variables, it will prove extremely useful, allowing us to quantify the net effect of the flow of a system through a volume for which border properties are known. In the following sections we are going to use this equation to assert four key physical principles ( 0.7) in order to analyze the flow of fluids: mass conservation; change of linear momentum; change of angular momentum; energy conservation. 3.3 Mass conservation Video: with sufficient skills (and lots of practice!), it is possible for a musician to produce an uninterrupted stream of air into an instrument while still continuing to breathe, a technique called circular breathing. Can you identify the different terms of eq. 3/5 as they apply to the clarinetist s mouth? by David Hernando Vitores (CC-by-sa) In this section, we focus on simply asserting that mass is conserved (eq. 0/22 p.19). Our study of the fluid s properties at the borders of the control volume is made by replacing variable B by mass m. Thus db dt becomes dm sys dt, which by definition is zero. In a similar fashion, b B/m = m/m = 1 and consequently the Reynolds transport theorem (3/4) becomes: dm sys dt the time change of the system s mass = 0 = d dt = 0 = CV ρ dv + the rate of change of mass inside the control volume + CS ρ ( V rel n) da (3/5) the net mass flow at the borders of the control volume This equation 3/5 is often called continuity equation. It allows us to compare the incoming and outgoing mass flows through the borders of the control volume. When the control volume has well-defined inlets and outlets through which the term ρ( V rel n) can be considered uniform (as for example in fig. 3.3), this equation reduces to: 0 = d { ρ dv + ρv A } { + ρv A } (3/6) dt CV out in = d { ρ dv + ρ V A } { ρ V A } dt CV out in = d ρ dv + {ṁ} {ṁ} (3/7) dt CV out in 64 In equation 3/6, the term ρv A at each inlet or outlet corresponds to the local mass flow ±ṁ (positive inwards, negative outwards) through the boundary.

65 Figure 3.3 A control volume for which the system s properties are uniform at each inlet and outlet. Here eq. 3/5 translates as 0 = d dt CV ρ dv + ρ 3 V 3 A 3 + ρ 2 V 2 A 2 ρ 1 V 1 A 1. Figure CC-0 o.c. With equation 3/7 we can see that when the flow is steady ( 0.8), the last two terms amount to zero, and the integral ρ dv (the total amount of mass CV in the control volume) does not change with time. 3.4 Change of linear momentum In this section, we apply Newton s second law: we assert that the variation of system s linear momentum is equal to the net force being applied to it (eq. 0/23 p.19). Our study of the fluid s properties at the control surface is carried out by replacing variable B by the quantity m V : momentum. Thus, db sys dt becomes d(m V sys ) dt, which is equal to F net, the vector sum of forces applied on the system as it transits the control volume. In a similar fashion, b B/m = V and equation 3/4, the Reynolds transport theorem, becomes: d(m V sys ) dt = F net = d ρ V dv + dt CV the rate of change the vector sum of = of linear momentum forces on the system within the control volume + CS ρ V ( V rel n) da (3/8) the net flow of linear momentum through the boundaries of the control volume When the control volume has well-defined inlets and outlets through which the term ρ V ( V rel n) can be considered uniform (fig. 3.3), this equation reduces to: F net = d { ρ V dv + (ρ V A) V } { (ρ V A) V } (3/9) dt CV out in = d {ṁ V ρ V dv + } {ṁ V } (3/10) dt CV out in 65

66 Figure 3.4 The same control volume as in fig Here, since the system s properties are uniform at each inlet and outlet, eq. 3/8 translates as F net = d dt CV ρ V dv + ρ 3 V 3 A 3 V 3 + ρ 2 V 2 A 2 V 2 ρ 1 V 1 A 1 V 1. Figure CC-0 o.c. Let us observe the four terms of equation 3/10 for a moment, for they are full of subtleties. First, we notice that even if the flow is steady (an therefore that net (ṁ) = 0 kg s 1 since d dt = 0), the last two terms do not necessarily cancel each other (i.e. it is possible that (ṁ V ) net 0). The reverse also applies: it is quite possible that F net 0 even if the net momentum flow through the boundaries is null (that is, even if (ṁ V ) net = 0). This would be the case, if ρ dv (the total amount of momentum CV within the borders of the control volume) varies with time. Walking forwards and backwards within a rowboat would cause such an effect. From equation 3/10 therefore, we read that two distinct phenomena can result in a net force on the system: Video: as a person walks, the deflection of the air passing around their body can be used to sustain the flight of a paper airplane (a walkalong glider). Can you figure out the momentum flow entering and leaving a control volume surrounding the glider? by Y:sciencetoymaker (styl) A difference between the values of ṁ V of the fluid at the entrance and exit of the control volume (caused, for example, by a deviation of the flow or by a mass flow imbalance); A change in time of the momentum m V within the control volume (for example, with the acceleration or the variation of the mass of the control volume). These two factors may cancel each other, so that the system may well be able to travel through the control volume without any net force being applied to it. 66

67 3.5 Change of angular momentum In this third spin on the Reynolds transport theorem, we assert that the change of the angular momentum of a system about a point X is equal to the net moment applied on the system about this point (eq. 0/24 p.19). Our study of the fluid s properties at the borders of the control volume is made by replacing the variable B by the angular momentum r Xm m V. Thus, db sys dt becomes d( r Xm m V sys ) dt, which is equal to M net, the vector sum of moments applied on the system about point X as it transits through the control volume. In a similar fashion, b B/m = r V and equation 3/4, the Reynolds transport theorem, becomes: d( r Xm m V ) sys dt = M net,x = d dt the sum of moments applied to the system = CV Video: pre-lecture briefing for this chapter, part 2/2 by o.c. (CC-by) r Xm ρ V dv+ r Xm ρ ( V rel n) V da CS (3/11) the rate of change of the angular momentum in the control volume in which r Xm is a vector giving the position of any mass m relative to point X. + the net flow of angular momentum through the control volume s boundaries When the control volume has well-defined inlets and outlets through which the term r Xm ρ ( V rel n) V can be considered uniform (fig. 3.5), this equation reduces to: M net,x = d { r Xm ρ V dv + rxm ṁ V } { rxm ṁ V } dt CV out in (3/12) Video: rocket landing gone wrong. Can you compute the moment exerted by the top thruster around the base of the rocket as it (unsuccessfully) attempts to compensate for the collapsed landing leg? by Y:SciNews (styl) Figure 3.5 A control volume for which the properties of the system are uniform at each inlet or outlet. Here the moment about point X is M net,x d dt CV r Xm ρ V dv + r 2 ṁ 2 V 2 r 1 ṁ 1 V 1. Figure CC-0 o.c. Equation 3/12 allows us to quantify, with relative ease, the moment exerted on a system based on inlet and outlet velocities of a control volume. 67

68 3.6 Energy conservation We conclude our frantic exploration of control volume analysis with the first principle of thermodynamics. We now simply assert that the change in the energy of a system can only be due to well-identified transfers (eq. 0/25 p.19). Our study of the fluid s properties at the borders of the control volume is made by replacing variable B by an amount of energy E sys. Now, attributed to three contributors: de sys dt de sys dt can be = Q net in + W shaft, net in + W pressure, net in (3/13) where Q net in is the net power transfered as heat; W shaft, net in is the net power added as work with a shaft; and W pressure, net in is the net power required to enter and leave the control volume. It follows that b B/m = E/m e; and e is broken down into where i is the specific internal energy (J kg 1 ); e k the specific kinetic energy (J kg 1 ); and e p the specific potential energy (J kg 1 ). e = i + e k + e p (3/14) Now, the Reynolds transport theorem (equation 3/4) becomes: de sys dt = Q net in + W shaft, net in + W pressure, net in = d ρ e dv + ρ e ( V rel n) da dt CV CS (3/15) When the control volume has well-defined inlets and outlets through which the term ρ e ( V rel n) can be considered uniform, this equation reduces to: Q net in + W shaft, net in + W pressure, net in = d dt Q netin + W shaft, net in = d dt = d dt in in CV CV ρ e dv + {ṁ e} {ṁ e} (3/16) out {ṁ(i ρ e dv + + ek + e p ) } out {ṁ(i + ek + e p ) } W pressure, net in { ṁ CV ρ e dv + out ( i V 2 + дz in { ( ṁ i + 1 )} 2 V 2 + дz )} + out { ṁ p } { ṁ p } ρ ρ in Making use of the concept of enthalpy defined as h i + p/ρ, we obtain: Q net in + W shaft, net in = d { ( ρ e dv + ṁ h + 1 )} { ( dt CV 2 V 2 + дz ṁ h + 1 )} 2 V 2 + дz out in (3/17) the rate of change the net power = of energy inside received by the system the control volume 68 + the net energy flow rate exiting the control volume boundaries

69 This equation 3/17 is particularly attractive, but it necessitates the input of a large amount of experimental data to provide useful results. It is indeed very difficult to predict how the terms i and p/ρ will change for a given flow process. For example, a pump with given power Q net in and W shaft, net in will generate large increases of terms p, V and z if it is efficient, or a large increase of terms i and 1/ρ if it is inefficient. This equation 3/17, sadly, does not allow us to quantify the net effect of shear and the extent of irreversibilities in a fluid flow. 3.7 The Bernoulli equation The Bernoulli equation has very little practical use for us; nevertheless it is so widely used that we have to dedicate a brief section to examining it. We will start from equation 3/17 and add five constraints: 1. Steady flow. Thus d dt ρ e dv = 0. CV In addition, ṁ has the same value at inlet and outlet; 2. Incompressible flow. Thus, ρ stays constant; 3. No heat or work transfer. Thus, both Q net in and W shaft, net in are zero; 4. No friction. Thus, the fluid energy i cannot increase due to an input from the control volume; 5. One-dimensional flow. Thus, our control volume has only one known entry and one known exit, all fluid particles move together with the same transit time, and the overall trajectory is already known. With these five restrictions, equation 3/17 simply becomes: { 0 = ṁ(i + p } { ρ + e k + e p ) ṁ(i + p } ρ + e k + e p ) out = ṁ [ (i + p 2 ρ V дz 2) (i + p 1 ρ + 1 ] 2 V дz 1) and we here obtain the Bernoulli equation: in p 1 ρ V дz 1 = p 2 ρ V дz 2 (3/18) This equation describes the properties of a fluid particle in a steady, incompressible, friction-less flow with no energy transfer. The Bernoulli equation can also be obtained starting from the linear momentum equation (eq. 3/8 p.65): F net = d ρ V dv + ρ V ( V rel n) da dt CV CS 69

70 When considering a fixed, infinitely short control volume along a known streamline s of the flow, this equation becomes: d F pressure + d F shear + d F gravity = d ρ V dv + ρva d V dt along a streamline, where the velocity V is aligned (by definition) with the streamline. Now, adding the restrictions of steady flow ( d/ dt = 0) and no friction ( d F shear = 0), we already obtain: CV d F pressure + d F gravity = ρva d V The projection of the net force due to gravity d F gravity on the streamline segment ds has norm d F gravity d s = дρa dz, while the net force due to pressure is aligned with the streamline and has norm df pressure,s = A dp. Along this streamline, we thus have the following scalar equation, which we integrate from points 1 to 2: A dp ρдa dz = ρva dv ρ dp 1 dp д dz = V dv ρ 2 1 д dz = 2 1 V dv The last obstacle is removed when we consider flows without heat or work transfer, where, therefore, the density ρ is constant. In this way, we arrive to equation. 3/18 again: p 1 ρ V дz 1 = p 2 ρ V дz 2 Let us insist on the incredibly frustrating restrictions brought by the five conditions above: 1. Steady flow. This constrains us to continuous flows with no transition effects, which is a reasonable limit; 2. Incompressible flow. We cannot use this equation to describe flow in compressors, turbines, diffusers, nozzles, nor in flows where M > 0,6. 3. No heat or work transfer. We cannot use this equation in a machine (e.g. in pumps, turbines, combustion chambers, coolers). 4. No friction. This is a tragic restriction! We cannot use this equation to describe a turbulent or viscous flow, e.g. near a wall or in a wake. 5. One-dimensional flow. This equation is only valid if we know precisely the trajectory of the fluid whose properties are being calculated. 70 Among these, the last is the most severe (and the most often forgotten): the Bernoulli equation does not allow us to predict the trajectory of fluid particles. Just like all of the other equations in this chapter, it requires a control volume with a known inlet and a known outlet.

71 3.8 Limits of integral analysis Integral analysis is an incredibly useful tool in fluid dynamics: in any given problem, it allows us to rapidly describe and calculate the main fluid phenomena at hand. The net force exerted on the fluid as it is deflected downwards by a helicopter, for example, can be calculated using just a loosely-drawn control volume and a single vector equation. As we progress through exercise sheet 3, however, the limits of this method slowly become apparent. They are twofold. First, we are confined to calculating the net effect of fluid flow. The net force, for example, encompasses the integral effect of all forces due to pressure, shear, and gravity applied on the fluid as it transits through the control volume. Integral analysis gives us absolutely no way of distinguishing between those sub-components. In order to do that (for example, to calculate which part of a pump s mechanical power is lost to internal viscous effects), we would need to look within the control volume. Second, all four of our equations in this chapter only work in one direction. The value db sys / dt of any finite integral cannot be used to find which function ρbv da was integrated over the control surface to obtain it. For example, there are an infinite number of velocity profiles which will result in a net force of 12 N. Knowing the net value of an integral, we cannot deduce the conditions which lead to it. In practice, this is a major limitation on the use of integral analysis, because it confines us to working with large swaths of experimental data gathered at the borders of our control volumes. From the wake below the helicopter, we deduce the net force; but the net force tells us nothing about the shape of the wake. Clearly, in order to overcome these limitations, we are going to need to open up the control volume, and look at the details of the flow within perhaps by dividing it into a myriad of sub-control volumes. This is what we set ourselves to in chapter 4, with a thundering and formidable methodology we shall call derivative analysis. 71

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73 Exercise sheet 3 (Analysis of existing flows) last edited May 4, 2018 These lecture notes are based on textbooks by White [13], Çengel & al.[16], and Munson & al.[18]. Except otherwise indicated, we assume that fluids are Newtonian, and that: ρ water = kg m 3 ; p atm. = 1 bar; ρ atm. = 1,225 kg m 3 ; T atm. = 11,3 C; µ atm. = 1, N s m 2 ; д = 9,81 m s 2. Air is modeled as a perfect gas (R air = 287 J K 1 kg 1 ; γ air = 1,4; c pair = J kg 1 K 1 ). Reynolds Transport Theorem: db sys = d dt dt CV ρb dv + CS ρb ( V rel n) da (3/4) Mass conservation: dm sys dt = 0 = d dt ρ dv + CV CS ρ ( V rel n) da (3/5) Change in linear momentum: d(m V sys ) = F net = d dt dt CV ρ V dv + CS ρ V ( V rel n) da (3/8) Change in angular momentum: d( r Xm m V ) sys dt = M net,x = d dt r Xm ρ V dv + CV CS r Xm ρ ( V rel n) V da (3/11) 3.1 Pipe bend CC-0 o.c. A pipe with diameter 30 mm has a bend with angle θ = 130, as shown in fig Water enters and leaves the pipe with the same speed V 1 = V 2 = 1,5 m s 1. The velocity distribution at both inlet and outlet is uniform. Figure 3.6 A pipe bend, through which water is flowing. We assume that the velocity distribution at inlet and outlet is identical. Figure CC-0 o.c. 73

74 1. What is the mass flow traveling through the pipe? 2. What is the force exerted by the pipe bend on the water? 3. Represent the force vector qualitatively (i.e. without numerical data). 4. What would be the new force if all of the speeds were doubled? 3.2 Water jet White [13] Ex3.9 A horizontal water jet hits a vertical wall and is split in two symmetrical vertical flows (fig. 3.7). The water nozzle has a 3 cm 2 cross-sectional area, and the water speed at the nozzle outlet is V jet = 20 m s 1. In this exercise, we consider the simplest possible case, with the effects of gravity and viscosity neglected. In that case, the water is perfectly split into two vertical jets of identical mass flow, the speed is identical everywhere, and the velocity distributions are uniform. 1. What is the velocity of each water flow as it leaves the wall? 2. What is the force exerted by the water on the wall? Now, the wall moves longitudinally in the same direction as the water jet, with a speed V wall = 15 m s What is the new force exerted by the water on the wall? 4. What is the velocity of each water jet as it leaves the wall, relative to the nozzle? 5. How would the force be modified if the volume flow was kept constant, but the diameter of the nozzle was reduced? (briefly justify your answer, e.g. in 30 words or less) 74 Figure 3.7 A water jet flowing out of a nozzle (left), and impacting a vertical wall on the right. Figure CC-0 o.c.

75 3.3 Exhaust gas deflector A deflector is used behind a stationary aircraft during ground testing of a jet engine (fig. 3.8). Figure 3.8 A mobile exhaust gas deflector, used to deflect hot jet engine exhaust gases upwards during ground tests. Figure CC-0 o.c. The deflector is fed with a horizontal air jet with a quasi-uniform velocity profile; the speed is V jet = 600 km h 1, temperature 400 C and the pressure is atmospheric. As the exhaust gases travel through the pipe, their heat losses are negligible. Gases are rejected with a 40 angle relative to the horizontal. The inlet diameter is 1 m and the horizontal outlet surface is 6 m What is the force exerted on the ground by the deflection of the exhaust gases? 2. Describe qualitatively (i.e. without numerical data) a modification to the deflector that would reduce the horizontal component of force. 3. What would the force be if the deflector traveled rearwards (positive x-direction) with a velocity of 10 m s 1? 3.4 Pelton water turbine White [13] P3.56 A water turbine is modeled as the following system: a water jet exiting a stationary nozzle hits a blade which is mounted on a rotor (fig. 3.9). In the ideal case, viscous effects can be neglected, and the water jet is deflected entirely with a 180 angle. Figure 3.9 Schematic drawing of a water turbine blade. This type of turbine is called Pelton turbine. Figure CC-0 o.c. 75

76 The nozzle has a cross-section diameter of 5 cm and produces a water jet with a speed V jet = 15 m s 1. The rotor diameter is 2 m and the blade height is negligibly small. We first study the case in which the rotor is prevented from rotating, so that the blade is stationary (V blade = 0 m s 1 ). 1. What is the force exerted by the water on the blade? 2. What is the moment exerted by the blade around the rotor axis? 3. What is the power transmitted to the rotor? We now let the rotor rotate freely. Friction losses are negligible, and it accelerates until it reaches maximum velocity. 4. What is the rotor rotation speed? 5. What is the power transmitted to the rotor? The rotor is now coupled to an electrical generator. 6. Show that the maximum power that can be transmitted to the generator occurs for V blade = 1 3 V water. 7. What is the maximum power that can be transmitted to the generator? 8. How would the above result change if viscous effects were taken into account? (briefly justify your answer, e.g. in 30 words or less) 3.5 Snow plow derived from Gerhart & Gross [6] Ex5.9 A road-based snow plow (fig. 3.10) is clearing up the snow on a flat surface. We wish to quantify the power required for its operation. The snow plow is advancing at 25 km h 1 ; its blade has a frontal-view width of 4 m. The snow on the ground is 30 cm deep and has density 300 kg m 3. The snow is pushed along the blade and is rejected horizontally with a 30 angle to the left of the plow. Its density has then risen to 450 kg m 3. The cross-section area A outlet of the outflowing snow in the x-y plane is 1,1 m What is the force exerted on the blade by the deflection of the snow? (Indicate its magnitude and coordinates) 2. What is the power required for the operation of the snow plow? 3. If the plow velocity was increased by 10 %, what would be the increase in power? 76

77 Figure 3.10 Outline schematic of a blade snow plow. 3.6 Drag on a cylindrical profile Figure CC-0 o.c. In order to measure the drag on a cylindrical profile, a cylindrical tube is positioned perpendicular to the air flow in a wind tunnel (fig. 3.11), and the longitudinal component of velocity is measured across the tunnel section. Figure 3.11 A cylinder profile set up in a wind tunnel, with the air flowing from left to right. Figure CC-0 o.c. 77

78 Upstream of the cylinder, the air flow velocity is uniform (u 1 = U = 30 m s 1 ). Downstream of the cylinder, the speed is measured across a 2 m height interval. Horizontal speed measurements are gathered and modeled with the following relationship: u 2(y) = 29 + y 2 (3/19) The width of the cylinder (perpendicular to the flow) is 2 m. The Mach number is very low, and the air density remains constant at ρ = 1,23 kg m 3 ; pressure is uniform all along the measurement field. What is the drag force applying on the cylinder? How would this value change if the flow in the cylinder wake was turbulent, and the function u 2(y) above only modeled time-averaged values of the horizontal velocity? (briefly justify your answer, e.g. in 30 words or less) 3.7 Drag on a flat plate We wish to measure the drag applying on a thin plate positioned parallel to an air stream. In order to achieve this, measurements of the horizontal velocity u are made around the plate (fig. 3.12). Figure 3.12 Side view of a plate positioned parallel to the flow. Figure CC-0 o.c. At the leading edge of the plate, the horizontal velocity of the air is uniform: u 1 = U = 10 m s 1. At the trailing edge of the plate, we observe that a thin layer of air has been slowed down by the effect of shear. This layer, called boundary layer, has a thickness of δ = 1 cm. The horizontal velocity profile can be modeled with the relation: u 2(y) = U ( y δ ) 1 7 (3/20) The width of the plate (perpendicular to the flow) is 30 cm and it has negligible thickness. The flow is incompressible (ρ = 1,23 kg m 3 ) and the pressure is uniform. 1. What is the drag force applying on the plate? 2. What is the power required to compensate the drag? Under which form is the kinetic energy lost by the flow carried away? Can this new form of energy be measured? (briefly justify your answer, e.g. in 30 words or less)

79 3.8 Drag measurements in a wind tunnel A group of students proceeds with speed measurements in a wind tunnel. The objective is to measure the drag applying on a wing profile positioned across the tunnel test section (fig. 3.13). Figure 3.13 Wing profile positioned across a wind tunnel. The horizontal velocity distributions upstream and downstream of the profile are also shown. Figure CC-0 o.c. Upstream of the profile, the air flow velocity is uniform (u 1 = U = 50 m s 1 ). Downstream of the profile, horizontal velocity measurements are made every 5 cm across the flow; the following results are obtained: vertical position (cm) horizontal speed u 2 (m s 1 ) The width of the profile (perpendicular to the flow) is 50 cm. The airflow is incompressible (ρ = 1,23 kg m 3 ) and the pressure is uniform across the measurement surface. 1. What is the drag applying on the profile? 2. How would the above calculation change if vertical speed measurements were also taken into account? 79

80 3.9 Moment on gas deflector We revisit the exhaust gas deflector of exercise 3.3 p.75. Figure 3.14 below shows the deflector viewed from the side. The midpoint of the inlet is 2 m above and 5 m behind the wheel labeled A, while the midpoint of the outlet is 3,5 m above and 1,5 m behind it. Figure 3.14 Side view of the mobile exhaust gas deflector which was shown in fig. 3.8 p.75 Figure CC-0 o.c. What is the moment generated by the gas flow about the axis of the wheel labeled A? 3.10 Helicopter tail moment In a helicopter, the role of the tail is to counter exactly the moment exerted by the main rotor about the main rotor axis. This is usually done using a tail rotor which is rotating around a horizontal axis. A helicopter, shown in figure 3.15, is designed to use a tail without a rotor, so as to reduce risks of accidents when landing and taking off. The tail is a long hollow cylindrical tube with two inlets and one outlet. To simplify calculations, we consider that pressure is atmospheric at every inlet and outlet. inlet A has a cross-section area of 0,2 m 2. It contributes hot exhaust gases of density 0,8 kg m 3 and velocity 12 m s 1, aligned with the (x) axis of the tail; inlet B contributes 25 kg s 1 of atmospheric air incoming at an angle α = 130 relative to the axis of the tail, with a velocity of 3 m s 1. The mix of exhaust gases and atmospheric air is rejected at the tip of the tail (outlet C) with a fixed velocity of 45 m s 1. The angle θ at which gases are rejected is controlled by the flight computer. 1. What is the rejection angle θ required so that the tail generates a moment of +6 kn m around the main rotor (y) axis? Propose and quantify a modification to the tail geometry or operating conditions that would allow the tail to produce no thrust (that is to say, zero force in the x-axis), while still generating the same moment.

81 Figure 3.15 Top-view of a helicopter using a rotor-less tail. Figure derived from a figure CC-by-sa by Commons User:FOX 52 Remark: this system is commercialized by MD Helicopters as the notar. The use of exhaust gases was abandoned, however, a clever use of air circulation around the tail pipe axis contributes to the generated moment; this effect is explored in chapter 8 ( p.176) Pressure losses in a pipe flow Non-examinable. Based on White [13] Water is circulated inside a cylindrical pipe with diameter 1 m (fig. 3.16). Figure 3.16 Velocity profiles at the inlet and outlet of a circular pipe. Figure CC-0 o.c. At the entrance of the pipe, the speed is uniform: u 1 = U av. = 5 m s 1. At the outlet of the pipe, the velocity profile is not uniform. It can be modeled as a function of the radius with the relationship: u 2(r ) = U center ( 1 r R ) 1 7 (3/21) 81

82 The momentum flow at the exit may be described solely in terms of the average velocity U av. with the help of a momentum flux correction factor β, such that ρ u ( ) 2 2 da = βρau av.. 2 When u 2 = U center 1 r m, R it can be shown that β = (1+m) 2 (2+m) 2 2(1+2m)(2+2m), and so with m = 1 7, we have β 1,02. [difficult question] What is the net force applied on the water so that it may travel through the pipe? 3.12 Thrust reverser non-examinable A turbofan installed on a civilian aircraft is equipped with a thrust inverter system. When the inverters are deployed, the cold outer flow of the engine is deflected outside of the engine nacelle. We accept that the following characteristics of the engine, when it is running at full power, and measured from a reference point moving with the engine, are independent of the aircraft speed and engine configuration: Inlet diameter D 1 = 2,5 m Inlet speed V 1 = 60 m s 1 Inlet temperature T 1 = 288 K Cold flow outlet velocity V 3 = 85 m s 1 Cold flow temperature T 3 = 300 K Hot flow outlet diameter D 4 = 0,5 m Hot flow outlet velocity V 4 = 200 m s 1 Hot flow outlet temperature T 4 = 400 K Air pressure at all exits p 3 = p 4 = p atm. = 1 bar Bypass ratio ṁ 3 /ṁ 4 = 6 Figure 3.17 Conceptual layout sketch of a turbofan. The airflow is from left to right. Figure CC-by o.c. We first study the engine in normal (forward thrust) configuration. 1. When mounted on a stationary test bench, what is the thrust provided by the engine? 2. In the approximation that the operating parameters remain unchanged, what is the thrust generated when the engine moves with a speed of 40 m s 1? We now deploy the thrust reversers. They deflect the cold flow with a 75 angle relative to the aircraft longitudinal axis. 3. What is the net thrust developed when the aircraft velocity is 40 m s 1? What is the net thrust developed when the aircraft is stationary?

83 Answers 3.1 1) ṁ = 1,0603 kg s 1 2) F net = ( 0,5681; 1,2184) N 3) The force is quadrupled ) F net on water = 120 N, and F water/wall = F net on water ; 3) F net on water = 7,5 N; 4) V exit, absolute = (15; 5); V exit, absolute = 15,8 m s ) F net x = 9,532 kn & F net y = +1,479 kn : F net = 9,646 kn; 3) F net 2 = 8,525 kn ) F net = 883,6 N; 2) M net X = F net R = 883,6 N m; 3) W rotor = 0 W; 4) ω = 143,2 rpm (F net = 0 N); 5) W rotor = 0 W again; 6) W rotor, max = 1,963 V blade, optimal = 1 3 V water jet ) ṁ = kg s 1 ; F net x = +10,07 kn, F net z = +12,63 kn (force on blade is opposite); 2) W = F net V plow = F netx V 1 = 69,94 kw 3.6 F netx = ρl ( ) S2 u 2 2 Uu 2 dy = 95,78 N. 3.7 F netx = ρl δ ( 0 u 2 (y) Uu ) (y) dy = 7, N : W drag = U F netx = 0,718 W. 3.8 F netx ρlσ y [( u 2 2 Uu 2 ) δy ] = 64,8 N. 3.9 Re-use ṁ = 67,76 kg s 1, V 1 = 166,7 m s 1, V 2 = 33,94 m s 1 from ex.3.3. From trigonometry, find R 2 V2 = 1,717 m. Then, plug in eq. 3/12 p.67: M net = 18,64 kn m in z-direction ) Work eq. 3/11 down to scalar equation (in y-direction), solve for θ: θ = 123,1. 2) There are multiple solutions which allow both moment and force equations to be solved at the same time. r C can be shortened, the flow in C can be split into forward and rearward components, or tilted downwards etc. Reductions in ṁ B or V C are also possible, but quantifying them requires solving both equations at once V center = 1,2245U ; F net = +393 N (positive!) 3.12 ṁ cold = 297 kg s 1, ṁ hot = 59,4 kg s 1 ; F cold flow, normal, bench & runway = +74,25 kn, F hot flow, normal & reverse, bench & runway = +8,316 kn; F cold flow, reverse, bench & runway = 24,35 kn and F hot flow, normal, bench & runway = +8,316 kn. Adding the net pressure force due to the (lossless) flow acceleration upstream of the inlet, we obtain, on the bench: F engine bench, normal = 93,26 kn, F engine bench, reverse = +5,344 kn; and on the runway: F engine runway, normal = 92,5 kn and F engine bench, reverse = +6,094 kn. 83

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85 Fluid Mechanics Chapter 4 Predicting fluid flow last edited September 23, Motivation Eulerian description of fluid flow Problem description Total time derivative Mass conservation Change of linear momentum The Cauchy equation The Navier-Stokes equation for incompressible flow The Bernoulli equation Change of angular momentum Energy conservation and increase in entropy CFD: the Navier-Stokes equations in practice Principle Two problems with CFD Exercises 103 These lecture notes are based on textbooks by White [13], Çengel & al.[16], and Munson & al.[18]. 4.1 Motivation In this chapter we assign ourselves the daunting task of describing the movement of fluids in an extensive manner. We wish to express formally, and calculate, the dynamics of fluids the velocity field as a function of time in any arbitrary situation. For this, we develop a methodology named derivateive analysis. Let us not shy away from the truth: the methods developed here are disproportionately complex in comparison to the solutions they allow us to derive by hand. Despite this, this chapter is extraordinarily important, for two reasons: Video: pre-lecture briefing for this chapter by o.c. (CC-by) derivative analysis allows us to formally describe and relate the key parameters that regulate fluid flow, and so is the key to developing an understanding of any fluid phenomenon, even when solutions cannot be derived; it is the backbone for computational fluid dynamics (cfd) in which approximate flow solutions are obtained using numerical procedures. 4.2 Eulerian description of fluid flow Problem description From now on, we wish to describe the velocity and pressure fields of a fluid with the highest possible resolution. For this, we aim to predict and describe the trajectory of fluid particles ( 0.2.2) as they travel. 85

86 Newton s second law allows us to quantify how the velocity vector of a particle varies with time. If we know all of the forces to which one particle is subjected, we can obtain a streak of velocity vectors V particle = (u,v,w) = f (x,y,z,t) as the particle moves through our area of interest. This is a description of the velocity of one particle; we obtain a function of time which depends on where and when the particle started its travel: V particle = f (x 0,y 0,z 0,t 0,t). This is the process used in solid mechanics when we wish to describe the movement of an object, for example, a single satellite in orbit. This kind of description, however, is poorly suited to the description of fluid flow, for three reasons: Firstly, in order to describe a given fluid flow (e.g. air flow around the rear-view mirror standing out of a car), we would need a large number of initial points (x 0,y 0,z 0,t 0 ), and we would then obtain as many trajectories V particle. It then becomes very difficult to describe to study and describe a problem that is local in space (e.g. the wake immediately behind the car mirror), because this requires finding out where the particles of interest originated, and accounting for the trajectories of each of them. Secondly, the concept of a fluid particle is not well-suited to the drawing of trajectories. Indeed, not only can particles strain indefinitely, but they can also diffuse into the surrounding particles, blurring and blending themselves one into another. Finally, the velocity and other properties of a given particle very strongly depend on the properties of the surrounding particles. We have to resolve simultaneously the movement equations of all of the particles. A space-based description of properties one in which we describe properties at a chosen fixed point of coordinates x point,y point, z point,t point is much more useful than a particle-based description which depends on departure points x 0,y 0,z 0,t 0. It is easier to determine the acceleration of a particle together with that of its current neighbors, than together with that of its initial (former) neighbors. What we are looking for, therefore, is a description of the velocity fields that is expressed in terms of a fixed observation point V point = (u,v,w) = f (x point,y point,z point,t), through which particles of many different origins may be passing. This is termed a Eulerian flow description, as opposed to the particle-based Lagrangian description. Grouping all of the point velocities in our flow study zone, we will obtain a velocity field V point that is a function of time Total time derivative Let us imagine a canal in which water is flowing at constant and uniform speed u = U canal (fig. 4.1). The temperature T water of the water is constant (in time), but not uniform (in space). We measure this temperature with a stationary probe, reading T probe = T water on the instrument. Even though the temperature T water is constant, when reading the value measured at the probe, temperature will be changing with time: 86 dt water dt = 0 dt probe dt = dt water dx u water

87 In the case where the water temperature is not simply heterogeneous, but also unsteady (varying in time with a rate T / t), then that local change in time will also affect the rate which is measured at the probe: dt water dt 0 dt probe dt = T water t + T water probe x u water (4/1) We keep in mind that the rate T / t can itself be a function of time and space; in equation 4/1, it is its value at the position of the probe and at the time of measurement which is taken into account. This line of thought can be generalized for three dimensions and for any property A of the fluid (including vector properties). The property A of one individual particle can vary as it is moving, so that it has a distribution A = f (x,y,z,t) within the fluid. The time rate change of A measured at a point fixed in space is named the total time derivative or simply total derivative 1 of A and written DA/Dt: D Dt t + u x + v y + w z DA Dt = A t + u A x + v A y + w A z D A Dt = A t + u A x + v A y + w A z = A x t A y t A z t + u A x x + u A y x + u A z x + v A x y + v A y y + v A z y + w A x z + w A y z + w A z z (4/2) (4/3) (4/4) (4/5) Figure 4.1 A one-dimensional water flow, for example in a canal. The water has a non-uniform temperature, which, even if it is constant in time, translates in a temperature rate change in time at the probe. Figure CC-0 o.c. 1 Unfortunately this term has many denominations across the litetature, including advective, convective, hydrodynamic, Lagrangian, particle, substantial, substantive, or Stokes derivative. In this document, the term total derivative is used. 87

88 In eq. equation 4/3, it is possible simplify the notation of the last three terms. We know the coordinates of the velocity vector V (by definition) and of the operator gradient (as defined in eq. 1/8 p. 32) are: V i u + j v + k w i x + j y + k z We thus define the advective operator, ( V ): V = u x + v y + w z (4/6) We can now rewrite eqs. 4/2 and 4/3 in a more concise way: D Dt t + ( V ) (4/7) DA Dt = A t + ( V )A (4/8) Video: half-century-old, but timeless didactic exploration of the concept of Lagrangian and Eulerian derivatives, with accompanying notes by Lumley[3] by the National Committee for Fluid Mechanics Films (ncfmf, 1969[12]) (styl) As we explored in equation 4/1, the total derivative of a property can be non-null even if the flow is perfectly steady; similarly, the total derivative of a property can be zero (e.g. when a stationary probe provides a constant measurement) even if the properties of the particles have a non-zero time change (e.g. a fluid with falling temperature, but with a non-uniform temperature). The total time derivative is the tool that we were looking for. Now, as we solve for the fluid properties in a stationary reference frame, instead of in the reference frame of a moving particle, we will replace all the time derivatives d dt with total derivatives D Dt. This allows us to compute the change in time of a property locally, without the need to track the movement of particles along our field of study. The first and most important such property we are instested in is velocity. Instead of solving for the acceleration of numerous particles ( d V particle / dt), we will focus on calculating the acceleration field D V /Dt. 4.3 Mass conservation The first physical principle we wish to express in a derivative manner is the conservation of mass (eq. 0/22 p.19). Let us consider a fluid particle of volume dv, at a given instant in time (fig. 4.2). We can reproduce our analysis from chapter 3 by quantifying the time change of mass within an arbitrary volume. In the present case, the control volume is stationary and the particle (our system) is flowing through it. We start with eq. 3/5 p.64: dm particle = 0 = d ρ dv + ρ( V rel n) da (4/9) dt dt CV CS 88

89 Figure 4.2 Conservation of mass within a fluid particle. In the x-direction, a mass flow ṁ 1 = ρ 1 u 1 dz dy is flowing in, and a mass flow ṁ 2 = ρ 2 u 2 dz dy is flowing out. These two flows may not be equal, since mass may also flow in the y- and z-directions. Figure CC-0 o.c. The first of these two integrals can be rewritten using the Leibniz integral rule: d ρ ρ dv = dt CV CV t dv + ρv S da CS ρ = dv (4/10) t CV where V S is the speed of the control volume wall; and where the term CS ρv S da is simply zero because we chose a fixed control volume, such as a fixed computation grid 1. Now we turn to the second term of equation 4/9, CS ρ( V rel n) da, which represents the mass flow ṁ net flowing through the control volume. In the direction x, the mass flow ṁ net x flowing through our control volume can be expressed as: ṁ net x = ρ 1 u 1 dz dy + ρ 2 u 2 dz dy CS CS = (ρu) dx dz dy CS x = (ρu) dv (4/11) x CV The same applies for directions y and z, so that we can write: ρ( V rel n) da = ṁ net = ṁ net x + ṁ net y + ṁ net z CS [ = CV x (ρu) + y (ρv) + ] z (ρw) dv = (ρ V ) dv (4/12) CV 1 In a cfd calculation in which the grid is deforming, this term CS ρv S da will have to be re-introduced in the continuity equation. 89

90 Now, with these two equations 4/10 and 4/12, we can come back to equation 4/9, which becomes: dm particle = 0 = d ρ dv + ρ( V rel n) da dt dt CV CS ρ 0 = t dv + (ρ V ) dv CV CV (4/13) Since we are only concerned with a very small volume dv, we drop the integrals, obtaining: for all flows, with all fluids. 0 = ρ t + (ρ V ) (4/14) This equation 4/14 is named continuity equation and is of crucial importance in fluid mechanics. All fluid flows, in all conditions and at all times, obey this law. Equation 4/14 is a scalar equation: it gives us no particular information about the orientation of V or about its change in time. In itself, this is insufficient to solve the majority of problems in fluid mechanics, and in practice it is used as a kinematic constraint to solutions used to evaluate their physicality or the quality of the approximations made to obtain them. The most assiduous readers will have no difficulty reading the following equation, ρ t + (ρ V ) = ρ t + V ρ + ρ( V ) = 0 (4/15) which allows us to re-write equation 4/14 like so: 1 ρ Dρ Dt + V = 0 (4/16) Therefore, we can see that for an incompressible flow, the divergent of velocity V (sometimes called volumetric dilatation rate) is zero: for any incompressible flow. V = 0 (4/17) In three Cartesian coordinates, this is re-expressed as: u x + v y + w z = 0 (4/18) 90

91 4.4 Change of linear momentum Now that we have seen what mass conservation tells us about the dynamics of fluids, we turn to the momentum equation, hoping to express the velocity field to the forces that are applied on the fluid The Cauchy equation We start by writing Newton s second law (eq. 0/23 p.19) as it applies to a fluid particle of mass m particle, as shown in fig Fundamentally, the forces on a fluid particle are of only three kinds, namely weight, pressure, and shear: 1 m particle d V dt = F weight + F net, pressure + F net, shear (4/19) We now apply this equation to a stationary cube of infinitesimal volume dv which is traversed by fluid. Measuring the velocity from the reference frame of the cube, and thus using a Eulerian description, we obtain: m particle dv D V Dt ρ D V Dt = m dv д + 1 dv F net, pressure + 1 dv F net, shear = ρ д + 1 dv F net, pressure + 1 dv F net, shear (4/20) Therefore, we will have obtained an expression for the acceleration field D V /Dt if we can find an expression for the net pressure and shear force. Let us quantify those efforts as they apply on each of the faces of the fluid Figure 4.3 In our study of fluid mechanics, we consider only forces due to gravity, shear, or pressure. Figure CC-0 o.c. 1 In some very special and rare applications, electromagnetic forces may also apply. 91

92 particle. For pressure, we obtain: F net, pressure,x = dy dz [p 1 p 4 ] [ = dy dz p ] x dx F net, pressure,x = dv p x F net, pressure,y = dv p y F net, pressure,z = dv p z (4/21) (4/22) (4/23) This pattern, we had seen in chapter 1 (with eq.1/8 p.32), can more elegantly be expressed with the gradient operator (see also Appendix A2 p.217), allowing us to write: 1 dv F net, pressure = p (4/24) As for shear, we can put our work from chapter 2 to good use now. There, we had seen that the effect of shear on a volume had eighteen components, the careful examination of which allowed us to quantify the net force due to shear (which has only three components). In the x-direction, for example, we had seen (with eq. 2/8 p.49, as illustrated again in fig. 4.4) that the net shear force in the x-direction F shearx was the result of shear in the x-direction on each of the six faces: F shear x = dx dy( τ zx 3 τ zx 6 ) + dx dz( τ yx 2 τ yx 5 ) + dz dy( τ xx 1 τ xx 4 ) This, in turn, was more elegantly expressed using the divergent operator (see also Appendix A2 p.217) as equation 2/13: ( τzx F shear x = dv z + τ yx y + τ ) xx (4/25) x = dv τ ix (4/26) Figure 4.4 Shear efforts on a cubic fluid particle (already represented in fig. 2.1 p.48). Figure CC-0 o.c. 92

93 Finally, we had assembled the components in all three directions into a single expression for the net shear force with eq. 2/14 p.50: F shear = dv τ ij (4/27) Now, we can put together our findings, eq. 4/24 for the force due to pressure and eq. 4/27 for the force due to shear, back into equation 4/20, we obtain the Cauchy equation: ρ D V Dt for all flows, with all fluids. = ρ д p + τ ij (4/28) The Cauchy equation is an expression of Newton s second law applied to a fluid particle. It expresses the time change of the velocity vector measured at a fixed point (the acceleration field D V /Dt) as a function of gravity, pressure and shear effects. This is quite a breakthrough. Within the chaos of an arbitrary flow, in which fluid particles are shoved, pressurized, squeezed, and distorted, we know precisely what we need to look for in order to quantify the time-change of velocity: gravity, the gradient of pressure, and the divergent of shear. Nevertheless, while it is an excellent start, this equation isn t detailed enough for us. In our search for the velocity field V, the changes in time and space of the shear tensor τ ij and pressure p are unknowns. Ideally, those two terms should be expressed solely as a function of the flow s other properties. Obtaining such an expression is what Claude-Louis Navier and Gabriel Stokes set themselves to in the 19 th century: we follow their footsteps in the next paragraphs The Navier-Stokes equation for incompressible flow The Navier-Stokes equation is the Cauchy equation (eq. 4/28) applied to Newtonian fluids. In Newtonian fluids, which we encountered in chapter 2 ( p.52), shear efforts are simply proportional to the rate of strain; thus, the shear component of eq. 4/28 can be re-expressed usefully. We had seen with eq. 2/16 p.51 that the norm τ ij of shear component in direction j along a surface perpendicular to i can be expressed as: τ ij = µ u j i Thus, in a Newtonian fluid, on each of the six faces of the infinitesimal cube from fig. 4.4, the shear in x-direction is proportional to the partial derivative of u (the component of velocity in the x-direction) with respect to the direction perpendicular to each face: τ xx1 = µ u x i 1 τ yx2 = µ u y i 2 τ zx3 = µ u z i 3 τ xx4 = µ u x i 4 τ yx5 = µ u y i 5 τ zx6 = µ u z i 6 93

94 These are the six constituents of the x-direction shear tensor τ ix. Their net effect is a vector expressed as the divergent τ ix, which, in incompressible flow 1, is expressed as: τ ix = τ xx x = ( µ u x i ) x = µ ( u x x = µ + τ yx y + τ zx z ) + ( µ u y i ) y ) i + µ ( u y + ( µ u z i ) z ) i + µ y ( 2 u ( x) u ( y) u ( z) 2 ( u z z i ) i (4/29) Using the Laplacian operator (see also Appendix A2 p.217) to represent the spatial variation of the spatial variation of an object: 2 (4/30) 2 A A (4/31) 2 A 2 A x 2 A y 2 A z = A x A y A z (4/32) we can re-write eq. 4/29 more elegantly and generalize to three dimensions: τ ix = µ 2 u i = µ 2 u (4/33) τ iy = µ 2 v j = µ 2 v (4/34) τ iz = µ 2 w k = µ 2 w (4/35) The three last equations are grouped together simply as τ ij = µ 2 V (4/36) With this new expression, we can come back to the Cauchy equation (eq. 4/28), in which we can replace the shear term with eq. 4/36. This produces the glorious Navier-Stokes equation for incompressible flow: ρ D V Dt for all incompressible flows of a Newtonian fluid. = ρ д p + µ 2 V (4/37) This formidable equation describes the property fields of all incompressible flows of Newtonian fluids. It expresses the acceleration field (left-hand side) as the sum of three contributions (right-hand side): those of gravity, gradient of pressure, and divergent of shear. The solutions we look for in equation 4/37 are the velocity (vector) field V = (u,v,w) = f 1 (x,y,z,t) and the pressure field p = f 2 (x,y,z,t), given a set of constraints to represent the problem at hand. These constraints, called boundary conditions, may be expressed in terms of 94 1 When the flow becomes compressible, normal stresses τ xx will see an additional term related to the change in density within the particle. An additional term, µ ( V ), would appear on the right side of eq. 4/37. This is well outside of the scope of this course.

95 velocities (e.g. the presence of a fixed solid body is expressed as a region for which V = 0) or pressure (e.g. a discharge into an open atmosphere may be expressed as a region of known constant pressure). The combination of equations 4/37 and 4/14 (continuity & Navier-Stokes) may not be enough to predict the solution of a given problem, most especially if large energy transfers take place within the flow. In that case, the addition of an energy equation and an expression of the second law of thermodynamics, both in a form suitable for derivative analysis, may be needed to provide as many equations as there are unknowns. Though it is without doubt charming, equation 4/37 should be remembered for what it is really: a three-dimensional system of coupled equations. In Cartesian coordinates this complexity is more apparent: ρ ρ ρ [ u t + u u x + v u y + w u z [ v t + u v x + v v y + w v ] z ] = ρд x p x + µ [ 2 u ( x) u ( y) u ( z) 2 ] (4/38) ] = ρд y p [ 2 y + µ v ( x) v ( y) v ( z) 2 [ w t + u w x + v w y + w w ] = ρд z p [ 2 z z + µ w ( x) w (4/39) ] ( y) w ( z) 2 (4/40) Today indeed, 150 years after it was first written, no general expression has been found for velocity or pressure fields that would solve this vector equation in the general case. Nevertheless, in this course we will use it directly: to understand and quantify the importance of key fluid flow parameters, in chapter 6; to find analytical solutions to flows in a few selected cases, in the other remaining chapters. After this course, the reader might also engage into computational fluid dynamics (cfd) a discipline entirely architectured around this equation, and to which it purposes to find solutions as fields of discrete values. As a finishing remark, we note that when the flow is strictly two-dimensional, the Navier-Stokes equation is considerably simplified, shrinking down to the system: [ u ρ t + u u x + v u ] = ρд x p [ 2 ] y x + µ u ( x) u ( y) 2 (4/41) [ v ρ t + u v x + v v ] = ρд y p [ 2 ] y y + µ v ( x) v ( y) 2 (4/42) Unfortunately, this simplification prevents us from accounting for turbulence, which is a strongly three-dimensional phenomenon The Bernoulli equation We had made clear in chapter 3 ( 3.7) that the Bernoulli equation was extraordinarily limited in scope, so much that starting from the more general 95

96 integral expressions for energy or momentum conservation was much safer in practice. As a reminder of this fact, and as an illustration of the bridges that can be built between integral and derivative analysis, it can be instructive to derive the Bernoulli equation directly from the Navier-Stokes equation. Let us start by following a particle along its path in an arbitrary flow, as displayed in fig The particle path is known (condition 5 in 3.7), but its speed V is not. We are now going to project every component of the Navier-Stokes equation (eq. 4/37) onto an infinitesimal portion of trajectory d s. Once all terms have been projected, the Navier-Stokes equation becomes simply a scalar equation: ρ V t + ρ( V ) V = ρ д p + µ 2 V ρ V t d s + ρ( V ) V d s = ρ д d s p d s + µ 2 V d s Because the velocity vector V of the particle, by definition, is always aligned with the path, its projection is always equal to its norm: V d s = V ds. Also, the downward gravity д and the upward altitude z have opposite signs, so that д d s = д dz; we thus obtain: ρ V t dv dp ds + ρ V ds = ρд dz ds ds ds + µ 2 V d s When we restrict ourselves to steady flow (condition 1 in 3.7), the first left-hand term vanishes. Neglecting losses to friction (condition 4) alleviates us from the last right-hand term, and we obtain: ρ dv dp V ds = ρд dz ds ds ds ρv dv = ρд dz dp This equation can then be integrated from point 1 to point 2 along the pathline: ρ V dv = ρд dz dp Figure 4.5 Different pathlines in an arbitrary flow. We follow one particle as it travels from point 1 to point 2. An infinitesimal path segment is named d s. Figure CC-0 o.c. 96

97 When no work or heat transfer occurs (condition 3) and the flow remains incompressible (condition 2), the density ρ remains constant, so that we indeed have returned to eq. 3/18: ( 1 ) 2 V 2 + д z + 1 p = 0 (4/43) ρ Thus, we can see that if we follow a particle along its path, in a steady, incompressible, frictionless flow with no heat or work transfer, its change in kinetic energy is due only to the result of gravity and pressure, in accordance with the Navier-Stokes equation. 4.5 Change of angular momentum There is no differential equation for conservation of angular momentum in fluid flow. Indeed, as the size of a considered particle tends to zero, it loses its ability to possess (or exchange) angular momentum about its center of gravity. Infinitesimal fluid particles are not able to rotate about themselves, only about an external point. So, in this chapter, no analysis is carried out using the concept of angular momentum. 4.6 Energy conservation and increase in entropy This topic is well covered in Anderson [8] The last two key principles used in fluid flow analysis can be written together in a differential equation similar to the Navier-Stokes equation. Once again, we start from the analysis of transfers on an infinitesimal control volume. We are going to relate three energy terms in the following form, naming them A, B and C for clarity: the rate of change of energy inside the fluid element = the net flux of heat into the element + the rate of work done on the element due to body and surface forces A = B + C (4/44) Let us first evaluate term C. The rate of work done on the particle is the dot product of its velocity V and the net force F net applying to it: C = V F weight F dv + net, pressure dv + F net, shear dv dv C = V (ρ д p + ) τ ij dv and for the case of a Newtonian fluid in an incompressible flow, this expression can be re-written as: C = V (ρ д p + µ 2 V ) dv (4/45) We now turn to term B, the net flux of heat into the element. We attribute this flux to two contributions, the first named Q radiation from the emission or absorption of radiation, and the second, named Q conduction to thermal 97

98 conduction through the faces of the element. We make no attempt to quantify Q radiation, simply stating that Q radiation = ρ q radiation dv (4/46) in which q radiation is the local power per unit mass (in W kg 1 ) transfered to the element, to be determined from the boundary conditions and flow temperature distribution. In the x-direction, thermal conduction through the faces of the element causes a net flow of heat Q conduction,x expressed as a function of the power per area q (in W m 2 ) through each of the two faces perpendicular to x: Q conduction,x = [ q x = q x x = q x x ( q x + q x x dx )] dy dz dx dy dz (4/47) dv (4/48) Summing contributions from all three directions, we obtain: Q conduction = Q conduction,x + Q conduction,y + Q [ conduction,z qx = x + q y y + q ] z dv (4/49) z So we finally obtain an expression for B: B = Q radiation + Q ( conduction [ qx = ρ q radiation x + q y y + q ]) z z dv (4/50) Finally, term A, the rate of change of energy inside the fluid element, can be expressed as a function of the specific kinetic energy e k and specific internal energy i: A = ρ D (i + 1 ) Dt 2 V 2 dv (4/51) We are therefore able to relate the properties of a fluid particle to the principle of energy conservation as follows: ρ D (i + 1 ) ( [ Dt 2 V 2 qx = ρ q radiation x + q y y + q ]) z + V (ρ д p z + µ 2 V ) (4/52) This is a scalar equation it only has one dimension, and involves the length of the velocity vector, V [ u 2 + v 2 + w 2] 1 2. In an interesting hack, we are able to incorporate an expression for the second principle of thermodynamics in this equation simply by expressing the fluxes q i as a function of the temperature gradients. Indeed, expressing a heat flux as the result of a difference in temperature according to the Fourier law, q i = k T i (4/53) 98

99 we constrain the direction of the heat fluxes and thus ensure that all dissipative terms resulting in temperature increases cannot be fed back into other energy terms, thus increasing the overall entropy. Equation 4/53 inserted into eq. 4/52 yields: ρ D (i + 1 ) ( [ Dt 2 V 2 2 ]) T = ρ q radiation + k ( x) T ( y) T ( z) 2 + V (ρ д p + µ 2 V ) Making use of a Laplacian operator, we come to: (i V 2 ) ρ D Dt = ( ρ q radiation + k 2 T ) + V (ρ д p + µ 2 V ) (4/54) This equation has several shortcomings, most importantly because the term q radiation is not expressed in terms of fluid properties, and because µ is typically not independent of the temperature T. Nevertheless, it in principle brings closure to our system of continuity and momentum equations, and these two influences may be either neglected, or modeled numerically. 4.7 CFD: the Navier-Stokes equations in practice This topic is well covered in Versteeg & Malalasekera [11] Principle In our analysis of fluid flow from a derivative perspective, our five physical principles from 0.7 have been condensed into three equations (often loosely referred together to as the Navier-Stokes equations). Out of these, the first two, for conservation of mass (4/17) and linear momentum (4/37) in incompressible flows, are often enough to characterize most free flows, and should in principle be enough to find the primary unknown, which is the velocity field V : 0 = V ρ D V = ρ д p Dt + µ 2 V We know of many individual analytical solutions to this mathematical problem: they apply to simple cases, and we shall describe several such flows in the upcoming chapters. However, we do not have one general solution: one that would encompass all of them. For example, in solid mechanics we have long understood that all pure free fall movements can be described with the solution x = x 0 + u 0 t and y = y 0 + v 0 t дt 2, regardless of the particularities of each fall. In fluid mechanics, even though our analysis was carried out in the same manner, we have yet to find such one general solution or even to prove that one exists at all. It is therefore tempting to attack the above pair of equations from the numerical side, with a computer algorithm. If one discretizes space and time in small increments δx, δy, δz and δt, we could re-express the x-component of eq. 4/37 as: 99

100 [ δu t ρ δt + u δu x δx + v δu y δy + w δu ] z = ρд x δp x δz δx + µ δ δu x δx δx x + δ δu y δy δy y + δ δu z δz z δz (4/55) If we start with a known (perhaps guessed) initial field for velocity and pressure, this equation 4/55 allows us to isolate and solve for δu t, and therefore predict what the u velocity field would look like after a time increment δt. The same can be done in the y- and z-directions. Repeating the process, we then proceed to the next time step and so on, marching in time, obtaining at every new time step the value of u, v and w at each position within our computation grid. This is the fundamental working principle for computational fluid dynamics (cfd) today Two problems with CFD There are two flaws with the process described above. The first flaw is easy to identify: while we may be able to start the calculation with a known pressure field, we do not have a mean to predict p (and thus δp/δx in eq. 4/55) at the next position in time (t + δt). This is not a surprise. Going back to the Cauchy equation (eq. 4/28), we notice that we took care of the divergent of shear τ ij, but not of the gradient of pressure p. We have neither an expression p = f ( V ), nor an equation that would give us a value for p/ t. Cfd software packages deal with this problem in the most uncomfortable way: by guessing a pressure field p (t+δt), calculating the new velocity field V (t+δt), and then correcting and re-iterating with the help of the continuity equation, until continuity errors are acceptably small. This scheme and its implementation are a great source of anguish amongst numerical fluid dynamicists. The second flaw appears once we consider the effect of grid coarseness. Every decrease in the size of the grid cell and in the length of the time step increases the total number of equations to be solved by the algorithm. Halving each of δx, δy, δz and δt multiplies the total number of equations by 16, so that soon enough the experimenter will wish to know what maximum (coarsest) grid size is appropriate or tolerable. Furthermore, in many practical cases, we may not even be interested in an exhaustive description of the velocity field, and just wish to obtain a general, coarse description of the fluid flow. Using a coarse grid and large time step, however, prevents us from resolving small variations in velocity: movements which are so small they fit in between grid points, or so short they occur in between time steps. Let us decompose the velocity field into two components: one is the average flow (u,v,w), and the other the instantaneous fluctuation flow (u,v,w ), too fine to be captured by our grid: u i u i + u i (4/56) u i 0 (4/57) 100

101 With the use of definition 4/56, we re-formulate equation 4/38 as follows: [ (u + u ) ρ t + (u + u ) (u + u ) x = ρд x (p + p ) x + (v + v ) (u + u ) + (w + w ) (u + ] u ) y z [ 2 (u + u ) + µ ( x) (u + u ) ( y) (u + u ] ) ( z) 2 Taking the average of this equation thus expressing the dynamics of the flow as we calculate them with a finite, coarse grid yields, after some intimidating but easily conquerable algebra: ρ [ u t + u u x + v u y + w u z = ρд x p x + µ ] + ρ u u u + v x y ] [ 2 u ( x) u ( y) u ( z) 2 + w u z (4/58) Equation 4/58 is the x-component of the Reynolds-averaged Navier-Stokes equation (rans). It shows that when one observes the flow in terms of the sum of an average and an instantaneous component, the dynamics cannot be expressed solely according to the average component. Comparing eqs. 4/58 and 4/38 we find that an additional term has appeared on the left side. This term, called the Reynolds stress, is often re-written as ρ u i u j / j; it is not zero since we observe in practice that the instantaneous fluctuations of velocity are strongly correlated. The difference between eqs. 4/58 and 4/38 can perhaps be expressed differently: the time-average of a real flow cannot be calculated by solving for the time-average velocities. Or, more bluntly: the average of the solution cannot be obtained with only the average of the flow. This is a tremendous burden in computational fluid dynamics, where limits on the available computational power prevent us in practice from solving for these fluctuations. In the overwhelming majority of computations, the Reynolds stress has to be approximated in bulk with schemes named turbulence models. Diving into the intricacies of cfd is beyond the scope of our study. Nevertheless, the remarks in this last section should hopefully hint at the fact that an understanding of the mathematical nature of the differential conservation equations is of great practical importance in fluid dynamics. It is for that reason that we shall dedicate the exercises of this chapter solely to playing with the mathematics of our two main equations. 101

102 102

103 Exercise sheet 4 (Differential analysis) last edited March 17, 2017 These lecture notes are based on textbooks by White [13], Çengel & al.[16], and Munson & al.[18]. Except otherwise indicated, we assume that fluids are Newtonian, and that: ρ water = kg m 3 ; p atm. = 1 bar; ρ atm. = 1,225 kg m 3 ; T atm. = 11,3 C; µ atm. = 1, N s m 2 ; д = 9,81 m s 2. Air is modeled as a perfect gas (R air = 287 J K 1 kg 1 ; γ air = 1,4; c pair = J kg 1 K 1 ). Continuity equation: 1 ρ Dρ Dt + V = 0 (4/16) Navier-Stokes equation for incompressible flow: ρ D V Dt = ρ д p + µ 2 V (4/37) 4.1 Revision questions For the continuity equation (eq. 4/16), and then for the incompressible Navier-Stokes equation (eq. 4/37), 1. Write out the equation in its fully-developed form in three Cartesian coordinates; 2. State in which flow conditions the equation applies. Also, in order to revise the notion of total (or substantial) derivative: 3. Describe a situation in which the total derivative of a property is non-zero although the fluid property is independent of time. 4. Describe a situation in which the total derivative of a property is zero although the time rate of change of this property is non-zero. 4.2 Acceleration field Çengel & al. [16] E4-3 A flow is described with the velocity field V = (0,5 + 0,8x) i + (1,5 0,8y) j (in si units). What is the acceleration measured by a probe positioned at (2; 2; 2) at t = 3 s? 103

104 4.3 Volumetric dilatation rate A flow is described by the following field (in si units): der. Munson & al. [18] 6.4 u = x 3 + y 2 + z v = xy + yz + z 3 w = 4x 2 z z What is the volumetric dilatation rate field? What is the value of this rate at {2;2;2}? 4.4 Incompressibility Çengel & al. [16] 9-28 Does the vector field V = (1,6 + 1,8x) i + (1,5 1,8y) j satisfy the continuity equation for two-dimensional incompressible flow? 4.5 Missing components Two flows are described by the following fields: Munson & al. [18] E6.2 + Çengel & al. [16] 9-4 u 1 = x 2 + y 2 + z 2 v 1 = xy + yz + z w 1 =? u 2 = ax 2 + by 2 + cz 2 v 2 =? w 2 = axz + byz 2 What must w 1 and v 2 be so that these flows be incompressible? 4.6 Another acceleration field White [13] E4.1 Given the velocity field V = (3t) i + (xz) j + (ty 2 ) k (si units), what is the acceleration field, and what is the value measured at {2;4;6} and t = 5 s? 4.7 Vortex A vortex is modeled with the following two-dimensional flow: y u = C x 2 + y 2 x v = C x 2 + y 2 Çengel & al. [16] 9.27 Verify that this field satisfies the continuity equation for incompressible flow. 104

105 4.8 Pressure fields Çengel & al. [16] E9-13, White [13] 4.32 & 4.34 We consider the four (separate and independent) incompressible flows below: V 1 = (ax + b) i + ( ay + cx) j V 2 = (2y) i + (8x) j V 3 = (ax + bt) i + (cx 2 + ey) j ( V 4 = U x ) y i U 0 L L j The influence of gravity is neglected on the first three fields. Does a function exist to describe the pressure field of each of these flows, and if so, what is it? 105

106 Answers 4.1 1) For continuity, use eqs. 4/2 and 1/8 in equation 4/16. For Navier-Stokes, see eqs. 4/38, 4/39 and 4/40 p. 95; 2) Read 4.3 p. 88 for continuity, and p.93 for Navier-Stokes; 3) and 4) see p. 86. D V 4.2 Dt = (0,4 + 0,64x) i + ( 1,2 + 0,64y) j. At the probe it takes the value 1,68 i + 0,08 j (length 1,682 m s 2 ). 4.3 V = x 2 + x z; thus at the probe it takes the value ( V ) probe = 4 s Apply equation 4/18 p.90 to V : the answer is yes ) Applying equation 4/18: w 1 = 3xz 1 2 z2 + f (x,y,t) ; 2) idem, v 2 = 3axy bzy 2 + f (x,z,t). 4.6 D V Dt = (3) i+(3z+y 2 x)t j+(y 2 +2xyzt) k. At the probe it takes the value 3 i+250 j+496 k. 4.7 Apply equation 4/18 to V to verify incompressibility. 4.8 Note: the constant (initial) value p 0 is sometimes implicitly written in the unknown functions f. 1) p = ρ [ abx a2 x 2 + bcy a2 y 2] + p 0 + f (t) ; ( p 2) p = ρ ( 8x 2 + 8y 2) + p 0 + f (t) ; 3) x y describe the [ pressure ( with) a mathematical ] function; U 2 4) p = ρ 0 L x + x2 2L + y2 2L д x x д y y + p 0 + f (t). ) y ( p ) x, thus we cannot 106

107 Fluid Mechanics Chapter 5 Pipe flow last edited October 30, Motivation Perfect flow in ducts Viscous laminar flow in ducts The entry zone Laminar viscous flow in a one-dimensional duct Viscous laminar flow in cylindrical pipes Principle Velocity profile 111 Velocity profile Quantification of losses Viscous turbulent flow in cylindrical pipes Predicting the occurrence of turbulence Characteristics of turbulent flow Velocity profile in turbulent pipe flow Pressure losses in turbulent pipe flow Exercises 119 These lecture notes are based on textbooks by White [13], Çengel & al.[16], and Munson & al.[18]. 5.1 Motivation In this chapter we focus on fluid flow within ducts and pipes. This topic allows us to explore several important phenomena with only very modest mathematical complexity. In particular, we are trying to answer two questions: 1. What does it take to describe fluid flow in ducts? 2. How can we quantify pressure losses in pipes and the power necessary to overcome them? Video: pre-lecture briefing for this chapter by o.c. (CC-by) Perfect flow in ducts We begin with the simplest possible ducted flow case: a purely hypothetical fully-inviscid, incompressible, steady fluid flow in a one-dimensional pipe. Since there are no viscous effects, the velocity profile across the duct remains uniform (flat) all along the flow (fig. 5.1). If the cross-sectional area A is changed, then the principle of mass conservation (eqs. 0/22, 3/5) is enough to allow us to compute the change in velocity u = V : in uniform pipe flow. ρv 1 A 1 = ρv 2 A 2 (5/1) 107

108 Figure 5.1 Inviscid fluid flow in a one-dimensional duct. In this purely hypothetical case, the velocity distribution is uniform across a cross-section of the duct. The average velocity and pressure change with cross-section area, but the total pressure p 0 = p ρv av. 2 remains constant. Figure CC-0 o.c. This information, in turn allows us to compute the pressure change between two sections of different areas by using the principle of energy conservation. We notice that the flow is so simple that the five conditions associated with the use of the Bernoulli equation ( 3.7) are fulfilled: the flow is steady, incompressible, one-dimensional, has known trajectory, and does not feature friction or energy transfer. A simple application of eq. 3/18 p. 69 yields: p 2 p 1 = 1 2 ρ [ V2 2 V ] ρд(z2 z 1 ) (5/2) in steady, incompressible, inviscid pipe flow without heat or work transfer. Thus, in this kind of flow, pressure increases everywhere the velocity decreases, and vice-versa. Another way of writing this equation is by stating that at constant altitude, the total or dynamic pressure p total p 0 p ρv 2 remains constant: at constant altitude, in laminar inviscid straight pipe flow. p 0 = cst. (5/3) We shall, of course, turn immediately to more realistic flows. With viscous effects, one key assumption of the Bernoulli equation breaks down, and additional pressure losses p friction will occur, which we would like to quantify. 108

109 5.3 Viscous laminar flow in ducts The entry zone Let us observe the velocity profile at the entrance of a symmetrical duct in a steady, laminar, viscous flow (fig. 5.2). Figure 5.2 Fluid flow velocity distributions at the entrance of a duct. All throughout the entrance region, the core region of the flow is accelerated, and the outer region decelerated, even though the flow itself is steady. Diagram CC-by-sa by Wikimedia Commons user:devender Kumar5908 The fluid enters the duct with a uniform (flat) velocity profile. Because of the no-slip condition at the wall ( p.50), the particles in contact with the duct walls are immediately stopped. Through viscosity, shear stress effects propagate progressively inwards. A layer appears in which viscosity effects are predominant, which we name boundary layer; this layer grows until it reaches the center of the duct. Past this point, the flow is entirely dictated by viscous effects and the velocity profile does not change with distance. The flow is then said to be fully developed Laminar viscous flow in a one-dimensional duct Once the flow is fully developed (but still steady and laminar), how can we describe the velocity and pressure distribution in our one-dimensional horizontal duct? We can start with a qualitative description. Because of the no-slip condition at the walls, the velocity distribution within any cross-flow section cannot be uniform (fig. 5.3). Shear occurs, which translates into a pressure decrease along the flow. The faster the flow, and the higher the gradient of velocity. Thus, shear within the flow, and the resulting pressure loss, both increase when the cross-sectional area is decreased. How can we now describe quantitatively the velocity profile and the pressure loss? We need a powerful, extensive tool to describe the flow and we turn to the Navier-Stokes equation which we derived in the previous chapter as eq. 4/37 p.94: Video: a young fluid dynamicist explores the mechanics of a granite ball fountain. The aesthetics are questionable but the physics are enjoyable (see exercise 5.4): a thin layer of water flowing in a laminar flow is enough to support a sphere weighing several tons. by World Travel (styl) ρ D V Dt = ρ д p + µ 2 V (5/4) Since we are applying this tool to the simple case of fully-developed, twodimensional incompressible fluid flow between two parallel plates (fig. 5.4), 109

110 Figure 5.3 Viscous laminar fluid flow in a one-dimensional pipe. This time, the no-slip condition at the wall creates a viscosity gradient across the duct cross-section. This in turn translates into pressure loss. Sudden duct geometry changes such as represented here would also disturb the flow further, but the effect was neglected here. Figure CC-0 o.c. we need only two Cartesian coordinates, reproducing eqs. 4/41 & 4/42: [ u ρ t + u u x + v u ] = ρд x p [ 2 ] y x + µ u ( x) u ( y) 2 (5/5) [ v ρ t + u v x + v v ] = ρд y p [ 2 ] y y + µ v ( x) v ( y) 2 (5/6) In this particular flow, we have restricted ourselves to a fully-steady ( d dt = 0), horizontal (д = д y ), one-directional flow (v = 0). When the flow is fully developed, u x = 0 and 2 u ( x) 2 = 0, and the system above shrinks down to: 0 = p [ 2 ] x + µ u ( y) 2 (5/7) 0 = ρд p y (5/8) Figure 5.4 Two-dimensional laminar flow between two plates, also called Couette flow. We already studied this flow case in fig. 2.4 p.55; this time, we wish to derive an expression for the velocity distribution. Figure CC-0 o.c. 110

111 We only have to integrate equation 5/7 twice with respect to y to come to the velocity profile across two plates separated by a height 2H: u = 1 ( ) p (y 2 H 2 ) (5/9) 2µ x Now, the longitudinal pressure gradient p x can be evaluated by working out the volume flow rate V for any given width Z with one further integration of equation 5/9: V Z = 2 H uz dy = 2H 3 ( ) p Z 0 3µ x p x = 3 µ 2 ZH 3 V (5/10) In this section, the overall process is more important than the result: by starting with the Navier-Stokes equations, and adding known constraints that describe the flow of interest, we can predict analytically all of the characteristics of a laminar flow. 5.4 Viscous laminar flow in cylindrical pipes We now turn to studying flow in cylindrical pipes, which are widely used; first considering laminar flow, and then expanding to turbulent flow Principle The process is identical to above, only applied to cylindrical instead of Cartesian coordinates. We focus on the fully-developed laminar flow of a fluid in a horizontal cylindrical pipe (fig. 5.5). For this flow, we wish to work out the velocity profile and calculate the pressure loss related to the flow. Figure 5.5 A cylindrical coordinate system to study laminar flow in a horizontal cylindrical duct. Figure CC-0 o.c. 111

112 5.4.2 Velocity profile We once again start from the Navier-Stokes vector equation, choosing this time to develop it using cylindrical coordinates: ρ v r t ρ ρ v r + v r r + v θ r = ρд r p r + µ [ vθ t v r θ v2 θ r + v z ( r v ) r r r [ 1 r v θ + v r r + v θ r = ρд θ 1 r [ vz t p θ + µ v z + v r r + v θ r = ρд z p z + µ v r z v θ θ + v ] rv θ v θ + v z r z [ ( 1 r v ) θ r r r ] v r θ + v v z z z ( r v ) z r r [ 1 r v r r r 2 2 v r ( θ ) 2 2 r 2 v θ θ + 2 v r ( z) 2 v θ r r 2 2 v θ ( θ ) r 2 v r θ + 2 v θ ( z) ] v z r 2 ( θ ) v z ( z) 2 ] (5/11) ] (5/12) (5/13) This mathematical arsenal does not frighten us, for the simplicity of the flow we are studying allows us to bring in numerous simplifications. Firstly, we have v r = 0 and v θ = 0 everywhere. Thus, by continuity, v z / z = 0. Furthermore, since our flow is symmetrical, v z is independent from θ. With these two conditions, the above system shrinks down to: 0 = ρд r p r 0 = ρд θ 1 p r θ 0 = p z + µ [ 1 r r ( r v )] z r (5/14) (5/15) (5/16) Equations (5/14) and (5/15) hold the key to describing pressure: p = ρдy + f (z) (5/17) this expresses the (unsurprising) fact that the pressure distribution within a vertical cross-section of the pipe is simply the result of a hydrostatic gradient. Now, with equation 5/16, we work towards obtaining an expression for v z by integrating twice our expression for v z / r: ( r v ) z = r p r r µ z ( r v ) z = r 2 ( ) p + k 1 r 2 µ z v z = r 2 ( ) p + k 1 ln r + k 2 (5/18) 4 µ z 112 We have to use boundary conditions so as to unburden ourselves from integration constants k 1 and k 2.

113 By setting v z (r=0) as finite, we deduce that k 1 = 0 (because ln(0) ). By setting v z (r=r) = 0 (no-slip condition), we obtain k 2 = R2 4 µ This simplifies eq. (5/18) and brings us to our objective, an extensive expression for the velocity profile across a pipe of radius R when the flow is laminar: v z = u (r ) = 1 4µ p z. ( ) p (R 2 r 2 ) (5/19) z This equation is parabolic (fig. 5.6). It tells us that in a pipe of given length L and radius R, a given velocity profile will be achieved which is a function only of the ratio p/µ. Figure 5.6 The velocity profile across a cylindrical pipe featuring laminar viscous flow. Figure CC-0 o.c. In the same way we proceeded before, we can express the pressure gradient in the pipe as a function of the volume flow rate, which itself is obtained through integration: V = p x = p friction L R 0 = 128 π ( ) v z (2πr ) dr = πd4 p 128 µ x µ V (5/20) D 4 This equation is interesting in several respects. For a given pipe length L and pressure drop p friction, the volume flow V increases with the power 4 of the diameter D. In other words, the volume flow is multiplied by 16 every time the diameter is doubled. We also notice that the pipe wall roughness does not appear in equation 5/20. In a laminar flow, increasing the pipe roughness has no effect on the velocity of the fluid layers at the center of the pipe. When the pipe is inclined relative to the horizontal with an angle α, eqs. (5/19) and (5/20) remain valid except for the pressure drop term p which is simply replaced by ( p friction + ρдl sin α), since gravity then contributes to the longitudinal change in pressure Quantification of losses Hydraulics is the oldest branch of fluid dynamics, and much of the notation used to describe pressure losses predates modern applications. The most widely-used parameters for quantifying losses due to friction in a duct are the following: 113

114 The elevation loss noted e is defined as e p friction ρд (5/21) It represents the hydrostatic height loss (with a positive number) caused by the fluid flow in the duct, and is measured in meters. It is often written h (without bars, but as a positive number), but in this document we reserve h to denote enthalpy (as in eq. 3/17 p.68). The Darcy friction factor noted f is defined as f p friction L 1 D 2 ρv av. 2 where V av. is the average flow velocity in the pipe. (5/22) This dimensionless term is widely used in the industry because it permits the quantification of losses independently of scale effects (L/D) and inflow conditions (ρv 2 av.). The loss coefficient noted K L is defined as K L p friction 1 2 ρv av. 2 (5/23) It is simply equivalent to a Darcy coefficient in which no value of D/L can be measured, and it is used mostly in order to describe losses in single components found in pipe systems, such as bends, filter screens, or valves. For a laminar flow in a cylindrical duct, for example, we can easily compute f, inserting equation 5/20 into the definition 5/22: f laminar cylinder flow = 32V av.µl L 1 D 2 ρv 2 in which we inserted the average velocity V av. = av.d = 64 µ 2 ρv av. D V p D2 = π R 2 32µL. (5/24) 5.5 Viscous turbulent flow in cylindrical pipes Predicting the occurrence of turbulence Video: very basic visualization of laminar and turbulent flow regimes in a transparent pipe by Engineering Fundamentals (styl) It has long been observed that pipe flow can have different regimes. In some conditions, the flow is unable to remain laminar (one-directional, fullysteady); it becomes turbulent. Although the flow is steady when it is averaged over short time period (e.g. a few seconds), it is subject to constant, smallscale, chaotic and spontaneous velocity field changes in all directions. In 1883, Osborne Reynolds published the results of a meticulous investigation into the conditions in which the flow is able, or not, to remain laminar (figs. 5.7 and 5.8). He showed that they could be predicted using a single non-dimensional parameter, later named Reynolds number, which, as we have seen already with eq. 0/27 p.21, is expressed as: [Re] ρ V L µ (5/25)

115 Figure 5.7 Illustration published by Reynolds in 1883 showing the installation he set up to investigate the onset of turbulence. Water flows from a transparent rectangular tank down into a transparent drain pipe, to the right of the picture. Colored die is injected at the center of the pipe inlet, allowing for the visualization of the flow regime. Image by Osborne Reynolds (1883, public domain) Figure 5.8 Illustration published by Reynolds in 1883 showing two different flow regimes observed in the installation from fig Image by Osborne Reynolds (1883, public domain) 115

116 In the case of pipe flow, the representative length L is conventionally set to the pipe diameter D and the velocity to the cross-section average velocity: [Re] D ρ V av. D µ (5/26) where V av. is the average velocity in the pipe (m s 1 ), and D is the pipe diameter (m). The occurrence of turbulence is very well documented (although it is still not well-predicted by purely-theoretical models). The following values are now widely accepted: Pipe flow is laminar for [Re] D ; Pipe flow is turbulent for [Re] D The significance of the Reynolds number extends far beyond pipe flow; we shall explore this in chapter 6 (pp.134 & 139) Characteristics of turbulent flow This topic is well covered in Tennekes & Lumley [4] Turbulence is a complex topic which is still not fully described analytically today. Although it may display steadiness when time-averaged, a turbulent flow is highly three-dimensional, unsteady, and chaotic in the sense that the description of its velocity field is carried out with statistical, instead of analytic, methods. In the scope of our study of fluid dynamics, the most important characteristics associated with turbulence are the following: A strong increase in mass and energy transfer within the flow. Slow and rapid fluid particles have much more interaction (especially momentum transfer) than within laminar flow; A strong increase in losses due to friction (typically by a factor 2). The increase in momentum exchange within the flow creates strong dissipation through viscous effects, and thus transfer (as heat) of macroscopic forms of energy (kinetic and pressure energy) into microscopic forms (internal energy, translating as temperature); Internal flow movements appear to be chaotic (though not merely random, as would be white noise), and we do not have mathematical tools to describe them analytically. Consequently, solving a turbulent flow requires taking account of flow in all three dimensions even for one-directional flow! Velocity profile in turbulent pipe flow In order to deal with the vastly-increased complexity of turbulent flow in pipes, we split each velocity component v i in two parts, a time-averaged component v i and an instantaneous fluctuation v i : v r = v r + v r 116 v θ = v θ + v θ v z = v z + v z

117 In our case, v r and v θ are both zero, but the fluctuations v r and v are not, and θ will cause v z to differ from the laminar flow case. The extent of turbulence is often measured with the concept of turbulence intensity I: I [ v 2 i ] 1 2 v i (5/27) Regrettably, we have not found a general analytical solution to turbulent pipe flow note that if we did, it would likely exhibit complexity in proportion to that of such flows. A widely-accepted average velocity profile constructed from experimental observations is: u (r ) = v z = v z max ( 1 r R ) 1 7 (5/28) While it closely and neatly matches experimental observations, this model is nowhere as potent as an analytical one and must be seen only as an approximation. For example, it does not allow us to predict internal energy dissipation rates (because it describes only time-averaged velocity), or even wall shear stress (because it yields ( u/ r ) r=r =, which is not physical). The following points summarize the most important characteristics of turbulent velocity profiles: They continuously fluctuate in time and we have no means to predict them extensively; They are much flatter than laminar profiles (fig. 5.9); They depend on the wall roughness; They result in shear and dissipation rates that are markedly higher than laminar profiles. Figure 5.9 Velocity profiles for laminar (A), and turbulent (B and C) flows in a cylindrical pipe. B represents the time-averaged velocity distribution, while C represents an arbitrary instantaneous distribution. Turbulent flow in a pipe also features velocities in the radial and angular directions, which are not shown here. Figure CC-0 o.c Pressure losses in turbulent pipe flow Losses caused by turbulent flow depend on the wall roughness ϵ and on the diameter-based Reynolds number [Re] D. For lack of an analytical solution, we are not able to predict the value of the friction factor f anymore. Several empirical models can be built to obtain f, 117

118 the most important of which is known as the Colebrook equation expressed as: 1 = 2 log 1 ϵ f 3,7 D + 2,51 [Re] D f (5/29) The structure of this equation makes it inconvenient to solve for f. To circumvent this difficulty, equation 5/29 can be solved graphically on the Moody diagram, fig This classic document allows us to obtain numerical values for f (and thus predict the pressure losses p friction ) if we know the diameter-based Reynolds number [Re] D and the relative roughness ϵ/d. 118 Figure 5.10 A Moody diagram, which presents values for f measured experimentally, as a function of the diameter-based Reynolds number [Re] D, for different values of the relative roughness ϵ/d. This figure is reproduced with a larger scale as figure 5.12 p.120. Diagram CC-by-sa S Beck and R Collins, University of Sheffield

119 Exercise sheet 5 (Pipe flow) last edited April 4, 2017 These lecture notes are based on textbooks by White [13], Çengel & al.[16], and Munson & al.[18]. Except otherwise indicated, we assume that fluids are Newtonian, and that: ρ water = kg m 3 ; patm. = 1 bar; ρ atm. = 1,225 kg m 3 ; Tatm. = 11,3 C; µ atm. = 1, N s m 2 ; д = 9,81 m s 2. Air is modeled as a perfect gas (R air = 287 J K 1 kg 1 ; γ air = 1,4; cpair = J kg 1 K 1 ). In cylindrical pipe flow, we accept the flow is always laminar for [Re]D , and always turbulent for [Re]D & The Darcy friction factor f is defined as: f p L1 2 D 2 ρvav. (5/22) The loss coefficient KL is defined as: KL p ρvav. (5/23) Viscosities of various fluids are given in fig Pressure losses in cylindrical pipes can be calculated with the help of the Moody diagram presented in fig p.120. Figure 5.11 Viscosity of various fluids at a pressure of 1 bar (in practice viscosity is almost independent of pressure). Figure White 2008 [13] 119

120 120 Figure 5.12 A Moody diagram, which presents values for f measured experimentally, as a function of the diameter-based Reynolds number [Re] D, for different relative roughness values. Diagram CC-by-sa S Beck and R Collins, University of Sheffield

121 5.1 Revision questions non-examinable The Moody diagram (fig p.120) is simple to use, yet it takes practice to understand it fully... here are three questions to guide your exploration. They can perhaps be answered as you work through the other examples. 1. Why is there no zero on the diagram? 2. Why are the curves sloped downwards should friction losses not instead increase with increasing Reynolds number? 3. Why can the pressure losses p be calculated given the volume flow V, but not the other way around? 5.2 Air flow in a small pipe Munson & al. [18] E8.5 A machine designed to assemble micro-components uses an air jet. This air is driven through a 10 cm-long cylindrical pipe with a 4 mm diameter, roughness 0,0025 mm, at an average speed of 50 m s 1. The inlet air pressure and temperature are 1,2 bar and 20 C; the air viscosity is quantified in fig p What is the pressure loss caused by the flow through the pipe? 2. Although that is not possible in practice, what would the loss in the case where laminar flow could be maintained throughout the pipe? 5.3 Couette flow CC-0 o.c. We consider the laminar flow of a fluid between two parallel plates (named Couette flow), as shown in fig Figure 5.13 Two-dimensional laminar flow between two plates, also called Couette flow. Figure CC-0 o.c. 1. Starting from the Navier-Stokes equations for incompressible, two-dimensional flow, [ u ρ t + u u x + v u ] = ρд x p [ 2 ] y x + µ u ( x) u ( y) 2 (4/41) [ v ρ t + u v x + v v ] = ρд y p [ 2 ] y y + µ v ( x) v ( y) 2 (4/42) 121

122 show that the velocity profile in a horizontal, laminar, fully-developed flow between two horizontal plates separated by a gap of height 2H is: u = 1 ( ) p (y 2 H 2 ) (5/9) 2µ x 2. Why would this equation fail to describe turbulent flow? 5.4 Kugel fountain Derived from Munson & al. [18] 6.91 A Kugel fountain is erected to entertain students of fluid mechanics (figs and 5.15). A granite block (2 700 kg m 3 ) is sculpted into a very smooth sphere of diameter 1,8 m. The sphere is laid on a cylindrical concrete stand of internal diameter 1,2 m, whose edges are also smoothed out. The stand is filled with water, and a pump creates a flow which lifts the sphere to make a decorative fountain. Water flows between the stand and the sphere along a length of 10 cm, with a thickness of 0,13 mm. Entrance effects are negligible, and because the Reynolds number is low, the flow is entirely laminar. 1. What is the pressure exerted by the sphere in the central part of the fountain? 2. The velocity distribution for flow in between two flat plates of width Z separated by an interval 2H, known as Couette flow, is modeled as a function of the stream-wise pressure gradient p/ x with the equation: u = 1 2µ ( p x ) (y 2 H 2 ) (5/9) Starting from this equation 5/9, show that the pressure gradient can be quantified as a function of the volume flow V: p x = 3 µ 2 ZH 3 V (5/10) 3. What is the pressure drop across the interval between the sphere and the concrete foundation? 4. What is the pumping power required for the fountain to work? 122 Figure 5.14 Schematic layout for a Kugel fountain. A pump (bottom left) increases the water pressure to a value high enough that it can support the weight of the sphere. A thin strip of water flows between the concrete foundation and the sphere. Figure CC-0 o.c.

123 5. How would this power change if the diameter of the concrete support (but not of the sphere) was increased? (briefly justify your answer, e.g. in 30 words or less) Figure 5.15 A Kugel fountain, featuring questionable aesthetics but providing the opportunity for amusing interaction. A granite sphere is supported by the flow of water between the sphere and a solid strip of concrete. Photo CC-by-sa by Commons User:Atamari 5.5 Water piping CC-0 o.c. A long pipe is installed to carry water from one large reservoir to another (fig. 5.16). The total length of the pipe is 10 km, its diameter is 0,5 m, and its roughness is ϵ = 0,5 mm. It must climb over a hill, so that the altitude changes along with distance. Figure 5.16 Layout of the water pipe. For clarity, the vertical scale is greatly exaggerated. The diameter of the pipe is also exaggerated. Figure CC-0 o.c. The pump must be powerful enough to push 1 m 3 s 1 of water at 20 C. Figure 5.11 p.119 quantifies the viscosity of various fluids, and fig p.120 quantifies losses in cylindrical pipes. 1. Will the flow in the water pipe be turbulent? 2. What is the pressure drop generated by the water flow? 3. What is the pumping power required to meet the design requirements? 4. What would be the power required for the same volume flow if the pipe diameter was doubled? 123

124 5.6 Pipeline CC-0 o.c. The Trans-Alaska Pipeline System is a cylindrical smooth steel duct with 1,22 m diameter, average roughness ϵ = 0,15 mm, and length km. Approximately 700 thousand barrels of oil ( m 3 ) transit through the pipe each day. The density of crude oil is approximately 900 kg m 3 and its viscosity is quantified in fig p.119. The average temperature of the oil during the transit is 60 C. In industrial pumps, oil starts to cavitate (change state, a very undesirable behavior) when its pressure falls below 0,7 bar. The pipeline is designed to withstand ground deformations due to seismic movements in several key zones, and crosses a mountain range with a total altitude variation of m. 1. How much time does an oil particle need to travel across the line? 2. Propose a pumping station arrangement, and calculate the power required for each pump. 3. How would the pumping power change if the speed was increased? 5.7 Pump with pipe expansion A pump is used to carry a volume flow of 200 L s 1 from one large water reservoir to another (fig. 5.17). The altitude of the water surface in both reservoirs is the same. Figure 5.17 Layout of the water pipe. For clarity, the diameter of the pipe and the vertical scale are exaggerated. Figure CC-0 o.c. The pipe connecting the reservoirs is made of concrete (ϵ = 0,25 mm); it has a diameter of 50 cm on the first half, and 100 cm on the second half. In the middle, the conical expansion element induces a loss coefficient of 0,8. At the outlet (at point D), the pressure is approximately equal to the corresponding hydrostatic pressure in the outlet tank. The inlet is 14 m below the surface. The total pipe length is 400 m; the altitude change between inlet and outlet is 12 m. 1. Represent qualitatively (that is to say, showing the main trends, but without displaying accurate values) the water pressure as a function of pipe distance, when the pump is turned off. 2. On the same graph, represent qualitatively the water pressure when the pump is switched on What is the water pressure at points A, B, C and D?

125 5.8 A more complex ducted flow non-examinable. From institute archives A more complicated laminar flow case can be studied with the following case. In the middle of a vertical container of width 2h filled with oil (fig. 5.18), a plate of width b (perpendicular to the plane) and length L with negligible thickness is sinking at constant velocity v p. Under the assumption that the flow is laminar everywhere, steady and fully developed, give an analytical expression for the friction force applied on the plate. Figure 5.18 The hypothetical case of a plate sinking vertically in an oil reservoir. 5.9 Wind tunnel non-examinable Describe the main characteristics of a wind tunnel that could be installed and operated in the room you are standing in. In order to do this: Start by proposing key characteristics for the test section; From these dimensions, draw approximately an air circuit to feed the test section (while attempting to minimize the size of the fan, whose cost increases exponentially with diameter). Quantify the static and stagnation pressures along the air circuit, by estimating the losses generated by wall shear and in the bends (you may use data from fig. 5.19); Quantify the minimum power required to generate your chosen test section flow characteristics. 125

126 Filter screen: η = 0,05 Figure 5.19 Loss coefficients K L (here noted η) generated by the use of various components within wind tunnel ducts. Figure Barlow, Rae & Pope 1999 [9] 5.10 Politically incorrect fluid mechanics non-examinable In spite of the advice of their instructor, a group of students attempts to apply fluid mechanics to incommendable activities. Their objective is to construct a drinking straw piping system that can mix a drink of vodka and tonic water in the correct proportions (fig. 5.20). They use a Strawz kit of connected drinking straws, two bottles, and a glass full of ice and liquid water to cool the mix. For simplicity, the following information is assumed about the setup: Vodka is modeled as 40 % pure alcohol (ethanol) with 60 % water by volume; Ethanol density is 0,8 kg m 3 ; Tonic water is modeled as pure water; The viscosity of water and alcohol mixes is described in table 5.1 (use the nearest relevant value); The pipe bends induce a loss coefficient factor K Lbend = 0,5 each; The pipe T-junction induces a loss coefficient factor K L = 0,3 in the line direction and 1 in the branching flow; 126 Figure 5.20 Conceptual sketch of a student experiment. Figure CC-0 o.c.

127 Percentage of alcohol by weight Viscosity in centipoise 0 1, , , , , , , , , , ,2 Table 5.1 Viscosity of a mix of ethanol and water at 20 C. data from Bingham, Trans. Chem. Soc, 1902, 81, 179. The pipe has inner diameter D = 3 mm and roughness η = 0,0025 mm. The students wish to obtain the correct mix: one quarter vodka, three quarters tonic water. For given levels of liquid in the bottles, is there a straw pipe network configuration that will yield the correct mix, and if so, what is it? 127

128 Answers 5.2 1) Calculating inlet density with the perfect gas model, [Re] D = (turbulent), a Moody diagram read gives f 0,029, so p friction = 1 292,6 Pa = 0,0129 bar. 2) If the flow were (magically) kept laminar, with equation 5/24, p friction = 200 Pa. 5.3 The structure is given in the derivation of equation 5/9 p. 111, and more details about the math are given in the derivation of the (very similar) equation 5/19 p p = F W sphere A basin 35,3 W. = 71,51 kpa. Then V = 0,494 L s 1 which enables us to obtain W pump = 5.5 p alt. = ρд( ) = 1,57 bar and p friction = 51,87 bar : W pump = 5,345 MW. 5.6 The total pumping power is W = 10 MW (!). Be careful not to cavitate the oil in the ascending sections, and not to burst the pipe in the descending sections! 5.7 Pressure losses to friction are Pa in the first half, +71 Pa in the throat, 117 Pa in the second half. Add hydrostatic pressure to obtain final result. 5.8 Calculate the velocity distribution in the same way as for equation 5/9 p.111. Evaluate the pressure gradient p y using mass conservation (total cross-section mass flow is zero). With the full velocity distribution, derivate v with respect to x to obtain F τ = ρsдh + 2µS h v p. 5.9 This is a fun (and quick) exercise, but the results strongly depend on your proposed design! Get help during office hours The author cannot remember which exercise you are referring to. 128

129 Fluid Mechanics Chapter 6 Scale effects last edited April 2, Motivation Scaling forces Force coefficients Power coefficient, and other coefficients Physical similarity Scaling flows The non-dimensional Navier-Stokes equation The flow parameters of Navier-Stokes Scaling flows Flow parameters as force ratios Acceleration vs. viscous forces: the Reynolds number Acceleration vs. gravity force: the Froude number Acceleration vs. elastic forces: the Mach number Other force ratios The Reynolds number in practice Exercises 141 These lecture notes are based on textbooks by White [13], Çengel & al.[16], and Munson & al.[18]. 6.1 Motivation In this chapter, we use the tools we derived in chapters 3 and 4 to analyze scale effects in fluid mechanics. This study should allow us to answer two questions: How can a flow be adequately reproduced in a smaller or larger scale? How do forces and powers change when the flow is scaled? Video: pre-lecture briefing for this chapter by o.c. (CC-by) Scaling forces Force coefficients Let us begin as a designer working for an automobile manufacturer. We wish to investigate the air flow around a full-size car, using a small model in a wind tunnel. The question we seek to answer is: if we succeed in reproducing the exact real flow in the wind tunnel, how large (or small) will the aerodynamic forces be on the model? For this, we wish to have a sense of how flow-induced forces scale when fluid flows are scaled. A look back on chapter 3, and in particular eq. 3/8 p.65, reminds us that for a steady flow through a given control volume, the net force induced on the fluid is expressed by: { F net = ρ V ( V rel n) da = (ρ V A) V } (6/1) CS net 129

130 This equation tells us that the norm of the force vector F net is directly related to the norm of the vector Σ ρ V A V. In the selection of a scale by which to measure F net, it is therefore sensible to include a term proportional to the the density ρ, a term proportional to the area A, and a term proportional to the square of velocity V. This scale of fluid-induced force is conventionally measured using the force coefficient C F : where F is the considered fluid-induced force (N); ρ is a reference fluid density (kg m 3 ); S is a reference surface area (m 2 ); and V is a reference velocity (m s 1 ). F C F 1 2 ρsv (6/2) 2 In effect, the force coefficient relates the magnitude of the force exerted by the fluid on an object (F) to the rate of flow of momentum towards the object (ρsv 2 = ṁv ). It is worth making a few remarks about this equation. First, it is important to realize that eq. 6/2 is a definition: while the choice of terms is guided by physical principles, it is not a physical law in itself, and there are no reasons to expect C F (which has no dimension, and thus no unit) to reach any particular value in any given case. The choice of terms is also worth commenting: F can be any fluid-induced force; generally we are interested in quantifying either the drag F D (the force component parallel to the freestream velocity) or the lift F L (force component perpendicular to the free-stream velocity); the reference area S is chosen arbitrarily. It matters only that this area grow and shrink in proportion to the studied flow case. In the automobile industry, it is customary to choose the vehicle frontal area as a reference, while in the aeronautical industry the top-view wing area is customarily used. The square of a convenient reference length L can also be chosen instead; the choice of reference velocity V and density ρ is also arbitrary, since both these properties may vary in time and space within the studied fluid flow case. It matters merely that the chosen reference values are representative of the case; typically the free-stream (faraway) conditions V and ρ are used; the term 1/2 in the denominator is a purely arbitrary and conventional value. Force coefficients are meaningful criteria to compare and relate what is going on in the wind tunnel and on the full-size object: in each case, we scale the measured force according to the relevant local flow conditions Power coefficient, and other coefficients 130 During our investigation of the flow using a small-scale model, we may be interested in measuring not only forces, but also other quantities such as power this would be an important parameter, for example, when studying a pump, an aircraft engine, or a wind turbine.

131 Going back once again to chapter 3, and in particular eq. 3/15 p.68, we recall that power gained or lost by a fluid flowing steadily through a control volume could be expressed as: de sys dt = CS { ρ e ( V rel n) da = (ρ V A)e } (6/3) in which the specific energy e grows proportionally to 1 2 V 2. Thus, the power gained or lost by the fluid is directly related to the magnitude of the scalar Σ ρ V AV 2. A meaningful scale for the power of a machine can therefore be the amount of energy in the fluid that is made available to it every second, a quantity that grows proportionally to ρav 3. This scale of fluid flow-related power is conventionally measured using the power coefficient C P : net W C P 1 2 ρsv (6/4) 3 where W is the power added to or subtracted from the fluid (W). Other quantities related to fluid flow, such as pressure loss or shear friction, are measured using similarly-defined coefficients. We have already used the pressure loss coefficient K L p / 1 2 ρv av. 2 in chapter 5 (eq. 5/23 p.114), and we shall soon use the shear coefficient c f in the coming chapter (eq. 7/5 p.151) Physical similarity In the paragraphs above, we have already hinted that under certain conditions, the force and power coefficients in experiments of differing scales may have the same value. This happens when we have attained physical similarity. Several levels of similarity may be achieved between two experiments: geometrical similarity (usually an obvious requirement), by which we ensure similarity of shape between the solid boundaries, which are then all related by a scale factor r; kinematic similarity, by which we ensure that the motion of all solid and fluid particles is reproduced to scale thus that for a given time scale ratio r t the relative positions are scaled by r, the relative velocities by r/r t and the relative accelerations by (r/r t ) 2 ; dynamic similarity, by which we ensure that not only the movement, but also the forces are all scaled by the same factor. In more complex cases, we may want to observe other types of similarity, such as thermal or chemical similarity (for which ratios of temperature or concentrations become important). The general rule that we follow is that the measurements in two experiments can be meaningfully compared if the experiments are physically similar. A direct consequence for our study of fluid mechanics is that in two different experiments, the force and power coefficients will be the same if the flows are physically similar. In the design of experiments in fluid mechanics, considerable effort is spent to attain at least partial physical similarity. Once this is done, measurements can 131

132 easily be extrapolated from one case to the other because all forces, velocities, powers, pressure differences etc. can be connected using coefficients. Video: practical application of scale effects: translating measurements made a 60 % size formula one car wind tunnel model into their real size race car values, as in exercise 6.4 For example, a flow case around a car A may be studied using a model B. If dynamic similarity is maintained (and that is by no means an easy task!), then the flow dynamics will be identical. The drag force F D B measured on the model can then be compared to the force F D A on the real car, using coefficients: since C FD B = C FD A we have: FD B ρas AVA2 F D A = C FD A ρas AVA2 = C FD B ρas AVA2 = ρ B S BVB 2 ρa S A VA + =* FD B 2 ρ S V B B B, by the Sauber F1 team (styl) It should be noted that even when perfect dynamic similarity is achieved, fluid-induced forces do not scale with other types of forces. For example, if the length of model B is 1/10th that of car A, then S B = 0,12S A and at identical fluid density and velocity, we would have F D B = 0,12 F D A. Nevertheless, the weight F W B of model B would not be one hundredth of the weight of car A. If the same materials were used to produce the model, weight would stay proportional to volume, not surface, and we would have F W B = 0,13 F W A. This has important consequences when weight has to be balanced by flowinduced forces such as aerodynamic lift on an airplane. It is also the reason why large birds such as condors or swans do not look like, and cannot fly as slowly as mosquitoes and bugs!1 6.3 Video: when their movement is filmed and then viewed sped-up, fog banks can appear to flow like water. What time-lapse rate is required to achieve physical similarity? Scaling flows We have just seen that when two flows are dynamically similar, their comparison is made very easy, as measurements made in one case can be transposed to the other with the use of coefficients. This naturally raises the question: how can we ensure that two flows are flows are dynamically similar? The answer comes from a careful re-writing of the Navier-Stokes equation: two flows are dynamically similar when all the relevant flow coefficients have the same value. by Simon Christen (stvl) The non-dimensional Navier-Stokes equation In this section we wish to obtain an expression for the Navier-Stokes equation for incompressible flow that allows for an easy comparison of its constituents. We start with the original equation, which we derived in chapter 4 as eq. 4/37 p.94: ρ V~ ~ V~ = ρ~ ~ + µ ~ 2V~ + ρ (V~ ) д p t (6/5) What we would now like to do is separate the geometry from the scalars in ~ as the multiple of this equation. This is akin to re-expressing each vector A its length A and a non-dimensional vector A~, which has the same direction ~ but only unit length. as A These ideas are beautifully and smartly explored in Tennekes [7, 14].

133 In order to achieve this, we introduce a series of non-dimensional physical terms, starting with non-dimensional time t, defined as time t multiplied by the frequency f at which the flow repeats itself: t f t (6/6) A flow with a very high frequency is highly unsteady, and the changes in time of the velocity field will be relatively important. On the other hand, flows with very low frequencies are quasi-steady. In all cases, as we observe the flow, non-dimensional time t progresses from 0 to 1, after which the solution is repeated. We then introduce non-dimensional velocity V, a unit vector field equal to the velocity vector field divided by its own (scalar field) length V : V V V (6/7) Pressure p is non-dimensionalized differently, since in fluid mechanics, it is the pressure changes across a field, not their absolute value, that influence the velocity field. For example, in eq. 6/5 p can be replaced by (p p ) (in which p can be any constant faraway pressure). Now, the pressure field p p is non-dimensionalized by dividing it by a reference pressure difference p 0 p, obtaining: p p p p 0 p (6/8) If p 0 is taken to be the maximum pressure in the studied field, then p is a scalar field whose values can only vary between 0 and 1. Non-dimensional gravity д is simply a unit vector equal to the gravity vector д divided by its own length д: д д д (6/9) And finally, we define a new operator, the non-dimensional del, L (6/10) which ensures that vector fields obtained with a non-dimensional gradient, and the scalar fields obtained with a non-dimensional divergent, are scaled by a reference length L. These new terms allow us to replace the constituents of equation 6/5 each by a non-dimensional unit term multiplied by a scalar term representing its length or value: t = t f V = V V p p = p (p 0 p ) д = д д = 1 L Now, inserting these in equation 6/5, and re-arranging, we obtain: 133

134 ρ ( V V ) 1 f t + ρ ( V V 1 L ) ( V V ) = ρд д 1 L [p (p 0 p ) + p ] + µ 2 L 2 ( V V ) ρv f V t + ρvv 1 ( V L ) V = ρд д 1 L [p (p 0 p )] + µv 1 L 2 2 V [ρv f ] V [ ρv 2 ] ( t + V L ) [ V = [ρд] д p0 p ] [ ] µv p + 2 V L L 2 (6/11) In equation 6/11, the terms in brackets each appear in front of non-dimensional (unit) vectors. These bracketed terms all have the same dimension, i.e. kg m 2 s 2. Multiplying each by L (of dimension m 2 s 2 kg 1 ), we obtain a ρv 2 purely non-dimensional equation: [ ] f L V V t + [1] ( V ) V = [ ] дl V 2 д [ ] p0 p ρv 2 [ ] p µ + ρv L 2 V (6/12) Equation 6/12 does not bring any information on top of the original incompressible Navier-Stokes equation (eq. 6/5). Instead, it merely separates it into two distinct parts. The first is a scalar field (purely numbers, and noted in brackets), which indicates the magnitude of the acceleration field. The other is a unit vector field (a field of oscillating vectors, all of length one, and noted with stars), which represents the geometry (direction) of the acceleration field. In this form, we can more easily observe and quantify the weight of the different terms relative to one another. This is the role of the flow parameters The flow parameters of Navier-Stokes From here on, we convene to call the terms written in brackets in eq. 6/12 flow parameters, and label them with the following definitions: [St] f L (6/13) V [Eu] p 0 p ρ V 2 (6/14) [Fr] V (6/15) д L [Re] ρ V L µ (6/16) This allows us to re-write eq. 6/12 as: [St] V t + [1] V [ 1 ] V = Fr 2 [ д [Eu] 1 ] p + 2 V Re And finally, we arrive at the simple, elegant and formidable non-dimensional incompressible Navier-Stokes equation expressed with flow parameters: 134

135 [St] V t + [1] V V = 1 [Fr] 2 д [Eu] p + 1 [Re] 2 V (6/17) Equation 6/17 is an incredibly potent tool in the study of fluid mechanics, for two reasons: 1. It allows us to quantify the relative weight of the different terms in a given flow. In this way, it serves as a compass, helping us determine which terms can safely be neglected in our attempts to find a particular flow solution, merely by quantifying the four parameters [St], [Eu], [Fr], and [Re]. 2. It allows us to obtain dynamic similarity between two flows at different scales (fig. 6.1). In order that a fluid flow be representative of another flow (for example in order to simulate airflow around an aircraft with a model in a wind tunnel), it must generate the same vector field V. In order to do this, we must generate an incoming flow with the same four parameters [St], [Eu], [Fr], and [Re]. Let us therefore take the time to explore the signification of these four parameters. The Strouhal number [St] f L V (eq. 6/13) quantifies the influence of unsteadiness effects over the acceleration field. It does this by evaluating the importance of the representative frequency f at which the flow pattern repeats itself. Very high frequencies denote highly unsteady flows. When the frequency is very low, [St] is very small; the flow is then quasi-steady and it can be solved at a given moment in time as if it was entirely steady. The Euler number [Eu] p 0 p (eq. 6/14) quantifies the influence of the ρ V 2 pressure gradient over the acceleration field. It does this by comparing the largest relative pressure p 0 p to the flow of momentum in the flow field. The greater [Eu] is, and the more the changes in the velocity field V are likely to be caused by pressure gradients rather than viscosity, convection or unsteadiness. The Froude number [Fr] V д L (eq. 6/15) quantifies the relative importance of gravity effects. In practice, gravity effects only play an important role in free surface flows, such as waves on the surface of a water reservoir. In most other cases, gravity contributes only to a hydrostatic effect, which has little influence over the velocity field. The Reynolds number [Re] ρ V L µ (eq. 6/16, also eqs. 0/27 p.21 & 5/25 p.114) quantifies the influence of viscosity over the acceleration field. It does this by comparing the magnitude of inertial effects (ρv L) to viscous effects (µ). When [Re] is very large, viscosity plays a negligible role and the velocity field is mostly dictated by the inertia of the fluid. We return to the significance of the Reynolds number in 6.5 below. 135

136 The Mach number [Ma] V /c (eq. 0/10 p.15) compares the flow speed V with that of the molecules within the fluid particles (the speed of sound c). [Ma] does not appear in equation 6/17 because from the start in chapter 4, we chose to restrict ourselves to non-compressible flows. If we hadn t, an additional term would exist on the right-hand side, µ ( V ), which is null when the divergent of velocity V is zero (which is always so at low to moderate flow speeds). We shall explore the significance of the Mach number in chapter 9. These five flow parameters should be thought of scalar fields within the studied flow domain: there is one distinct Reynolds number, one Mach number etc. for each point in space and time. Nevertheless, when describing fluid flows, the custom is to choose for each parameter a single representative value for the whole flow. For example, when describing pipe flow, it is customary to quantify a representative Reynolds number [Re] D based on the average flow velocity V av. and the pipe diameter D (as we have seen with eq. 5/26 p. 116), while the representative Reynolds number for flow over an aircraft wing is often based on the free-stream velocity V and the wing chord length. Similarly, the flight Mach number [Ma] displayed on an aircraft cockpit instrument is computed using the relative free-stream air speed V and the free-stream speed of sound c, rather than particular values measured closer to the aircraft Scaling flows Although the above mathematical treatment may be intimidating, it leads us to a rather simple and useful conclusion, which can be summarized as follows. When studying one particular fluid flow around or within an object, it may be useful to construct a model that allows for less expensive or more comfortable measurements (fig. 6.1). In that case, we need to make sure that the fluid flow on the model is representative of the real case. This can be achieved only if the relevant flow parameters have the same value in both cases. If this can be achieved although in practice this can rarely be achieved exactly! then the force and power coefficients also keep the same value. The scale effects are then fully-controlled and the model measurements are easily extrapolated to the real case. Figure 6.1 In order for the flow around a wind tunnel model to be representative of the flow around the real-size aircraft (here, a 48 m-wide Lockheed C-141 Starlifter), dynamic similarity must be obtained. The value of all flow parameters must be kept identical. This is not always feasible in practice. Wind tunnel photo by NASA (public domain) Full-size aircraft photo CC-by by Peter Long 136

137 6.4 Flow parameters as force ratios This topic is well covered in Massey [5] Instead of the above mathematical approach, the concept of flow parameter can be approached by comparing forces in fluid flows. Fundamentally, understanding the movement of fluids requires applying Newton s second law of motion: the sum of forces which act upon a fluid particle is equal to its mass times its acceleration. We have done this in an aggregated manner with integral analysis (in chapter 3, eq. 3/8 p.65), and then in a precise and all-encompassing way with differential analysis (in chapter 4, eq. 4/37 p.94). With the latter method, we obtain complex mathematics suitable for numerical implementation, but it remains difficult to obtain rapidly a quantitative measure for what is happening in any given flow. In order to obtain this, an engineer or scientist can use force ratios. This involves comparing the magnitude of a type of force (pressure, viscous, gravity) either with another type of force, or with the mass-times-acceleration which a fluid particle is subjected to as it travels. We are not interested in the absolute value of the resulting ratios, but rather, in having a measure of the parameters that influence them, and being able to compare them across experiments Acceleration vs. viscous forces: the Reynolds number The net sum of forces acting on a particle is equal to its mass times its acceleration. If a representative length for the particle is L, the particle mass grows proportionally to the product of its density ρ and its volume L 3. Meanwhile, its acceleration relates how much its velocity V will change over a time interval t: it may be expressed as a ratio V / t. In turn, the time interval t may be expressed as the representative length L divided by the velocity V, so that the acceleration may be represented as proportional to the ratio V V /L. Thus we obtain: net force = mass acceleration ρl 3V V L F net ρl 2 V V We now observe the viscous force acting on a particle: it is proportional to the the shear effort and a representative acting surface L 2. The shear can be modeled as proportional to the viscosity µ and the rate of strain, which will grow proportionally to V /L. We thus obtain a crude measure for the magnitude of the shear force: viscous force µ V L L2 F viscous µ V L The magnitude of the viscous force can now be compared to the net force: net force viscous force ρl2 V V µ V L = ρv L µ = [Re] (6/18) 137

138 and we recognize the ratio as the Reynolds number (6/16). We thus see that the Reynolds number can be interpreted as the inverse of the influence of viscosity. The larger [Re] is, and the smaller the influence of the viscous forces will be on the trajectory of fluid particles Acceleration vs. gravity force: the Froude number The weight of a fluid particle is equal to its mass, which grows with ρl 3, multiplied by gravity д: weight force = F W ρl 3 д The magnitude of this force can now be compared to the net force: net force weight force ρl2 V 2 ρl 3 д = V 2 Lд = [Fr]2 (6/19) and here we recognize the square of the Froude number (6/15). We thus see that the Froude number can be interpreted as the inverse of the influence of weight on the flow. The larger [Fr] is, and the smaller the influence of gravity will be on the trajectory of fluid particles Acceleration vs. elastic forces: the Mach number In some flows called compressible flows the fluid can perform work on itself, and and the fluid particles then store and retrieve energy in the form of changes in their own volume. In such cases, fluid particles are subject to an elastic force in addition to the other forces. We can model the pressure resulting from this force as proportional to the bulk modulus of elasticity K of the fluid (formally defined as K ρ p/ ρ); the elastic force can therefore be modeled as proportional to KL 2 : elasticity force = F elastic KL 2 The magnitude of this force can now be compared to the net force: net force elasticity force ρl2 V 2 KL 2 = ρv 2 K This ratio is known as the Cauchy number; it is not immediately useful because the value of K in a given fluid varies considerably not only according to temperature, but also according to the type of compression undergone by the fluid: for example, it grows strongly during brutal compressions. During isentropic compressions and expansions (isentropic meaning that the process is fully reversible, i.e. without losses to friction, and adiabatic, i.e. without heat transfer), we will show in chapter 9 (with eq. 9/7 p.195) that the bulk modulus of elasticity is proportional to the square of the speed of sound c: K reversible = c 2 ρ (6/20) 138

139 The Cauchy number calibrated for isentropic evolutions is then net force ρv 2 elasticity force reversible K = V 2 c 2 = [Ma]2 (6/21) and here we recognize the square of the Mach number (0/10). We thus see that the Mach number can be interpreted as the influence of elasticity on the flow. The larger [Ma] is, and the smaller the influence of elastic forces will be on the trajectory of fluid particles Other force ratios The same method can be applied to reach the definitions for the Strouhal and Euler numbers given in 6.3. Other numbers can also be used which relate forces that we have ignored in our study of fluid mechanics. For example, the relative importance of surface tension forces or of electromagnetic forces are quantified using similarly-constructed flow parameters. In some applications featuring rotative motion, such as flows in centrifugal pumps or planetary-scale atmospheric weather, it may be convenient to apply Newton s second law in a rotating reference frame. This results in the appearance of new reference-frame forces, such as the Coriolis or centrifugal forces; their influence can then be studied using additional flow parameters. In none of those cases can flow parameters give enough information to predict solutions. They do, however, provide quantitative data to indicate which forces are relevant in which places: this not only helps us understand the mechanisms at work, but also distinguish the negligible from the influential, a key characteristic of efficient scientific and engineering work. 6.5 The Reynolds number in practice Among the five non-dimensional parameters described above, the Reynolds number [Re] is by far the most relevant in the study of most fluid flows, and it deserves a few additional remarks. As we have seen, the Reynolds number is a measure of how little effect the viscosity has on the time-change of the velocity vector field: With low [Re], the viscosity µ plays an overwhelmingly large role, and the velocity of fluid particles is largely determined by that of their own neighbors; With high [Re], the momentum ρv of the fluid particles plays a more important role than the viscosity µ, and the inertia of fluid particles affects their trajectory much more than the velocity of their neighbors. Video: half-century-old, but timeless didactic exploration how the dynamics of fluids change with Reynolds number, with accompanying notes by Taylor[2] by the National Committee for Fluid Mechanics Films (ncfmf, 1967[12]) (styl) In turn, this gives the Reynolds number a new role in characterization of flows: it can be thought of as the likeliness of the flow being turbulent over the length L. Indeed, from a kinematic point of view, viscous effects are highly stabilizing: they tend to harmonize the velocity field and smooth out disturbances. On the contrary, when these effects are overwhelmed by inertial effects, velocity non-uniformities have much less opportunity to dissipate, and at the slightest disturbance the flow will become and remain turbulent (fig. 6.2). This is the reason why the quantification of a representative Reynolds number is often the first step taken in the study of a fluid flow. 139

140 Figure 6.2 A viscous opaque fluid is dropped into a clearer receiving static fluid with identical viscosity. The image shows four different experiments photographed after the same amount of time has elapsed. The viscosity is decreased from left to right, yielding Reynolds numbers of 0,05, 10, 200 and respectively. As described in eq. 6/18, low Reynolds number indicate that viscous effects dominate the change in time of the flow field. As the Reynolds number increases, the nature of the velocity field changes until it becomes clearly turbulent. By the National Committee for Fluid Mechanics Films (ncfmf, 1967[12]), with accompanying notes by Taylor[2] Image Educational Development Center, Inc., reproduced under Fair Use doctrine A screen capture from film Low Reynolds Number Flow at 140

141 Exercise sheet 6 (Scale effects) last edited April 2, 2018 Except otherwise indicated, we assume that fluids are Newtonian, and that: ρ water = kg m 3 ; patm. = 1 bar; ρ atm. = 1,225 kg m 3 ; Tatm. = 11,3 C; µ atm. = 1, N s m 2 ; д = 9,81 m s 2. Air is modeled as a perfect gas (R air = 287 J K 1 kg 1 ; γ air = 1,4; cpair = J kg 1 K 1 ). Non-dimensional incompressible Navier-Stokes equation: [St] V~ ~ V~ = 1 д~ [Eu] ~ p + 1 ~ 2V~ + [1] V~ 2 [Re] t [Fr] in which [St] f L V, [Eu] p0 p, ρ V2 [Fr] V д L and [Re] (6/17) ρv L µ. The force coefficient C F and power coefficient C P are defined as: CF F ρsv CP W ρsv The speed of sound c in air is modeled as: p c = γ RT (6/4) (0/11) Figure 6.3 quantifies the viscosity of various fluids as a function of temperature. Figure 6.3 Viscosity of various fluids at a pressure of 1 bar (in practice viscosity is almost independent of pressure). Figure repeated from fig p.119; White 2008 [13] 141

142 6.1 Scaling with the Reynolds number CC-0 o.c. 1. Give the definition of the Reynolds number, indicating the si units for each term. 2. What is the consequence on the velocity field of having a low Reynolds number? A high Reynolds number? 3. Give an example of a high Reynolds number flow, and of a low Reynolds number flow. The standard golf ball has a diameter of 42,67 mm and a mass of 45,93 g. A typical maximum velocity for such balls is 200 km h 1. A student wishes to investigate the flow over a golf ball using a model in a wind tunnel. S/he prints an enlarged model with a diameter of 50 cm. 4. If the atmospheric conditions are identical, what flow velocity needs to be generated during the experiment in order to reproduce the flow patterns around the real ball? 5. Would the Mach number for the real ball then be reproduced? 6. Would this velocity need to be adjusted if the experiment was run on a very hot summer day? 7. If the model was made of the same material as the real ball, how heavy would it be? 6.2 Fluid mechanics of a giant airliner non-examinable, CC-0 o.c. In this exercise, we consider that that viscosity is entirely independent from pressure and density (this is a reasonable first approximation). An Airbus A380 airliner has an 80 m-wingspan and is designed to cruise at [Ma] = 0,85 where the air temperature and pressure are 40 C and 0,25 bar. A group of students wishes to study the flow field around the aircraft, with a wind tunnel whose test section has a diameter of 1 m (in which, obviously, the model has to fit!). 1. If the temperature inside the tunnel is maintained at T tunnel = 20 C, propose a combination of wind tunnel velocity and pressure which would enable the team to adequately model the effects of viscosity. 2. Which flow conditions would be required in a water tunnel with the same dimensions? Shocked by their answers to the above questions, the group of students decides instead to study the effect of compressibility (while accepting that viscous effects may not be adequately modeled) If the maximum velocity attainable in the wind tunnel is 80 m s 1, which air temperature is required for compressibility effects to be modeled?

143 6.3 Scale effects on a dragonfly A dragonfly (sketched in fig. 6.4) has a 10 cm wingspan, a mass of 80 mg, and cruises at 4 m s 1, beating its four wings 20 times per second. Figure 6.4 Plan view of a dragonfly Figure by o.c., A. Plank, Drury, Dru, Westwood, J. O. (CC-0) You are tasked with the investigation of the flight performance of the dragonfly, and have access to a wind tunnel with a test section of diameter 1 m. 1. Which model size and flow velocity would you use for this experiment? 2. How many wing beats per second would then be required on the model? 3. What would be the lift developed by the model during the experiment? 4. How much mechanical power would the model require, compared to the real dragonfly? 6.4 Formula One testing You are leading the Aerodynamics team in a successful Formula One racing team. Your team races a car that is 5,1 m long, 1,8 m wide, has a 610 kw power plant, and a mass of 750 kg. The car can reach a top speed of 310 km h 1, at which speed the drag force is 7,1 kn. In order to test different aerodynamic configurations for the car, your team invests in a 50-million euro wind tunnel in which you will run tests on a model. You currently have the choice between two possible model sizes: a 60 % model and a (smaller) 50 % model. 1. What would be the frontal area of each model, in proportion to the frontal area of the real car? 2. What would be the volume of each model, compared to the volume of the real car? 3. How much less weight would the 50 % model have than the 60 % model? In the end, the regulations for the Formula One racing change, and your team is forced to opt for a 50 % car model. 143

144 4. If the ambient atmospheric conditions cannot be changed, which flow speed in the wind tunnel is required, so that the air flow around the real car is reproduced around the model? In practice the regulations change once again, and the maximum speed authorized in wind tunnel tests is now 50 m s Your team considers modifying the air temperature to compensate for the limit in the air speed. If the temperature in the tunnel can be controlled between 10 C and 40 C, but the pressure remains atmospheric, what is the maximum race-track speed that can be reproduced in the wind tunnel? 6. In that case, by which factor should the model drag force measurements be multiplied in order to correspond to the real car? 7. [non-examinable question] The wind tunnel has a 2 m by 2 m square test crosssection. What is the power required to bring atmospheric air (T atm. = 15 C, V atm. = 0 m s 1 ) to the desired speed and temperature in the test section? NB: this exercise is inspired by an informative and entertaining video by the Sauber F1 team about their wind tunnel testing, which the reader is encouraged to watch at 144

145 Answers 6.1 1) See equation 6/16 and subsequent comments; 2) See 6.5, in practice this can be summarized in two or three sentences; 3) High-Re: air flow around an airliner in cruise; Low-Re: air flow around a dust particle falling to the ground. 4) Match [Re]: V 2 = 4,74 m s 1 5) No, but it is very low anyway (no compressibility effects to be reproduced) 6) Yes! (discuss in class) 7) m 2 = 73,9 kg 6.2 1) With a half-aircraft model of half-length L 2 2 = 80 cm, at identical speed, we would need an air density ρ 2 = 24,9 kg m 3. At ambient temperature, this requires a pressure p 2 = 20,9 bar! 2) In water, the density cannot be reasonably controlled, and we need a velocity V 3 = 418 m s 1! 3) In air, at V 4 = 80 m s 1, Mach number can be reproduced at T 4 = 251 C (although the Reynolds number is off). Wind tunnels used to investigate compressible flow around aircraft have very powerful coolers ) With e.g. a model of span 60 cm, match the Reynolds number: V 2 = 0,67 m s 1 ; 2) Match the Strouhal number: f 2 = 0,56 Hz. Mach, Froude and Euler numbers will have no effect here; 3) and 4) are left as a surprise for the student ) The third model would have 42 % less mass than the second; 4) Maintaining [Re] requires V 3 = 172,2 m s 1 (fast!); 5) Now the fastest track speed that can be studied is 27 m s 1 (assuming identical viscosity everywhere); 6) Multiply force measurements by 1,74 to scale up to reality; 7) One needs powerful coolers! 145

146 146

147 Fluid Mechanics Chapter 7 Flow near walls last edited October 30, Motivation Describing the boundary layer Concept Why do we study the boundary layer? Characterization of the boundary layer The laminar boundary layer Governing equations Blasius solution Pohlhausen s model Transition The turbulent boundary layer Separation Principle Separation according to Pohlhausen Exercises 161 These lecture notes are based on textbooks by White [13], Çengel & al.[16], and Munson & al.[18]. 7.1 Motivation In this chapter, we focus on fluid flow close to solid walls. In these regions, viscous effects dominate the dynamics of fluids. This study should allow us to answer two questions: How can we quantify shear-induced friction on solid walls? How can we describe and predict flow separation? Video: pre-lecture briefing for this chapter by o.c. (CC-by) Describing the boundary layer Concept At the very beginning of the 20 th century, Ludwig Prandtl observed that for most ordinary fluid flows, viscous effects played almost no role outside of a very small layer of fluid along solid surfaces. In this area, shear between the zero-velocity solid wall and the outer flow dominates the flow structure. He named this zone the boundary layer. We indeed observe that around any solid object within a fluid flow, there exists a thin zone which is significantly slowed down because of the object s presence. This deceleration can be visualized by measuring the velocity profile (fig. 7.1). The boundary layer is a concept, a thin invisible layer whose upper limit (termed δ, as we will see in 7.2.3) is defined as the distance where the fluid velocity in direction parallel to the wall is 99 % of the outer (undisturbed) flow velocity. 147

148 Figure 7.1 A typical velocity profile in a boundary layer. Only the horizontal component of velocity (V x = u) is represented. Figure CC-by Olivier Cleynen Upon investigation, we observe that the boundary layer thickness depends strongly on the main flow characteristics. In particular, it decreases when speed increases or when viscosity is decreased (fig. 7.2). As we travel downstream along a boundary layer, we observe experimentally that the flow regime is always laminar at first. Then, at some distance downstream which we name transition point, the boundary layer becomes turbulent. The flow-wise position of the transition point depends on the flow properties and is somewhat predictable. Further downstream, the boundary layer becomes fully turbulent. It has larger thickness than in the laminar regime, and grows at a faster rate. Like all turbulent flows, it then features strong energy dissipation and its analytical description becomes much more difficult. The flow within the boundary layer, and the main external flow (outside of it) affect one another, but may be very different in nature. A laminar boundary layer may exist within a turbulent main flow; while turbulent boundary layers are commonplace in laminar flows. 148 Figure 7.2 The thickness of the boundary layer depends strongly on the faraway incoming flow velocity U. Figure CC-by-sa Olivier Cleynen & Commons User:F l a n k e r

149 7.2.2 Why do we study the boundary layer? Expending our energy on solving such a minuscule area of the flow may seem counter-productive, yet three great stakes are at play here: First, a good description allows us to avoid having to solve the Navier-Stokes equations in the whole flow. Indeed, outside of the boundary layer and of the wake areas, viscous effects can be safely neglected. Fluid flow can then be described with ρ D V Dt = p (the Euler equation, which we will introduce as eq. 8/3 in the next chapter) with acceptably low inaccuracy. As we shall soon see, this allows us to find many interesting solutions, all within reach of human comprehension and easily obtained with computers. Unfortunately, they cannot account for shear on walls. Thus, solving the flow within the boundary layer, whether analytically or experimentally, allows us to solve the rest of the flow in a simplified manner (fig. 7.3). Figure 7.3 Fluid flow around a wing profile. When analyzing the flow, whether analytically or within a computational fluid dynamics (cfd) simulation, the flow domain is frequently split into three distinct areas. In the boundary layer (B), fluid flow is dominated by viscosity. Outside of the boundary layer (A), viscous effects are very small, and the flow can be approximately solved using the Euler equation. Lastly, in the turbulent wake (C), characterization of the flow is very difficult and typically necessitates experimental investigations. Figure CC-by-sa Olivier Cleynen Secondly, the boundary layer is the key to quantifying friction. A good resolution of the boundary layer allows us to precisely quantify the shear forces generated by a fluid on an object. Finally, a good understanding of the mechanisms at hand within the boundary layer allows us to predict flow separation, which is the divergence of streamlines relative to the object. Control of the boundary layer is key to ensuring that a flow will follow a desired trajectory! Characterization of the boundary layer Three different parameters are typically used to quantify how thick a boundary layer is at any given position. 149

150 The first is the thickness δ, δ y u=0,99u (7/1) which is, as we have seen above, equal to the distance away from the wall where the speed u is 99 % of U. The second is the displacement thickness δ, which corresponds to the vertical distance by which streamlines outside of the boundary layer have been displaced. This vertical shifting of the flow occurs because the inner fluid is slowed down near the wall, creating some blockage for the outer flow, which then proceeds to avoid it partially by deviating outwards (fig. 7.4). Integral analysis performed on a control volume enclosing the boundary layer (as for example in exercise 3.7 from chapter 3) allows us to quantify the displacement thickness as: δ 0 ( 1 u ) U dy (7/2) In practice, this integral can be calculated on a finite interval (instead of using the limit), as long as the upper limit exceeds the boundary layer thickness. The third and last parameter is the momentum thickness θ, which is equal to the thickness of a corresponding layer of fluid at velocity U which would possess the same amount of momentum as the boundary layer. The momentum thickness can be thought of as the thickness of the fluid that would need to be entirely stopped (for example by pumping it outside of the main flow) in order to generate the same drag as the local boundary layer. A review of the concepts of chapter 3 (e.g. in exercises 3.6 to 3.8) allows us to to quantify the momentum thickness as: θ δ 0 u U ( 1 u ) U dy (7/3) with the same remark regarding the upper limit. The momentum thickness, in particular when compared to the displacement thickness, is a key parameter used in prediction models for boundary layer separation. Once these three thicknesses have been quantified, we are generally looking for a quantification of the shear term τ wall. Since we are working with the hypothesis that the fluid is Newtonian, we merely have to know u (y) to Figure 7.4 Thickness δ and displacement thickness δ of a boundary layer. It is important to understand that the boundary layer is not a closed zone: streamlines (drawn blue) penetrate it and the vertical velocity v, although very small compared to u, is not zero. Figure CC-by-sa Olivier Cleynen & Commons User:F l a n k e r 150

151 quantify shear according to equation 2/16 (p.51): τ wall yx = µ u y (7/4) this expression is a function of x, and typically τ wall decreases with increasing distance inside a laminar boundary layer. The wall shear exerted by the boundary layer is typically non-dimensionalized with the shear coefficient c f, where U is the outer-layer (free-stream) velocity. c f(x ) τ wall 1 2 ρu 2 (7/5) The shear coefficient, just like the shear, remains a function of the flow-wise distance x. Once these parameters are known, the wall shear force is simply obtained by integrating τ over the entire surface S: F shearyx = τ wall yx dx dz (7/6) 7.3 The laminar boundary layer Governing equations S What is happening inside a laminar, steady boundary layer? We begin by writing out the Navier-Stokes for incompressible isothermal flow in two Cartesian coordinates (eqs. 4/41 & 4/42 p.95): [ u ρ t + u u x + v u ] = ρд x p [ 2 ] y x + µ u ( x) u ( y) 2 (7/7) [ v ρ t + u v x + v v ] = ρд y p [ 2 ] y y + µ v ( x) v ( y) 2 (7/8) Building from these two equations, we are going to add three simplifications, which are hypotheses based on experimental observation of fluid flow in boundary layers: 1. Gravity plays a negligible role; 2. The component of velocity perpendicular to the wall (in our convention, v) is very small (v u). Thus, its stream-wise spatial variations can also be neglected: v x 0 and 2 v 0. The same goes for the derivatives in the y-direction: ( x) 2 v y 0 and 2 v 0. ( y) 2 3. The component of velocity parallel to the wall (in our convention, u) varies much more strongly in the y-direction than in the x-direction: 2 u 2 u. ( x) 2 ( y) 2 151

152 With all of these simplifications, equation 7/8 shrinks down to p y 0 (7/9) which tells us that pressure is a function of x only ( p x = dp dx ). We now turn to equation 7/7, first to obtain an expression for pressure by applying it outside of the boundary layer where u = U : dp dx = ρu du dx and secondly to obtain an expression for the velocity profile: (7/10) u u x + v u y = 1 p ρ x + µ ρ 2 u ( y) 2 = U du dx + µ 2 u ρ ( y) 2 (7/11) Thus, the velocity field V = (u;v) = f (x,y) in a steady laminar boundary layer is driven by the two following equations, answering to Newton s second law and continuity: u u x + v u y = U du 2 u ( y) 2 (7/12) dx + µ ρ u x + v y = 0 (7/13) The main unknown in this system is the longitudinal speed profile across the layer, u (x,y). Unfortunately, over a century after it has been written, we still have not found an analytical solution to it Blasius solution Heinrich Blasius undertook a PhD thesis under the guidance of Prandtl, in which he focused on the characterization of laminar boundary layers. As part of his work, he showed that the geometry of the velocity profile (i.e. the velocity distribution) within such a layer is always the same, and that regardless of the flow velocity or the position, u can be simply expressed as a function of non-dimensionalized distance away from the wall termed η: ρu η y (7/14) µx Blasius was able to show that u is a function such that u U = f (η), with f f f = 0 unfortunately, no known analytical solution to this equation is known. However, it has now long been possible to obtain numerical values for f at selected positions η. Those are plotted in fig Based on this work, it can be shown that for a laminar boundary layer flowing along a smooth wall, the four parameters about which we are interested are 152

153 Figure 7.5 The velocity profile obtained by Blasius (an exact solution to the Navier- Stokes equations simplified with laminar boundary-layer hypothesis). Figure CC-0 o.c. solely function of the distance-based Reynolds number [Re] x : δ x = 4,91 [Re]x δ x = 1,72 [Re]x θ x = 0,664 [Re]x c f(x ) = 0,664 [Re]x (7/14a) (7/14b) (7/14c) (7/14d) Pohlhausen s model Contrary to Blasius who was searching for an exact solution, Ernst Pohlhausen made an attempt at modeling the velocity profile in the boundary layer with a simple equation. He imagined an equation of the form: u U = д (Y ) = ay + by 2 + cy 3 + dy 4 (7/15) in which Y y/δ represents the distance away from the wall, and he then set to find values for the four coefficients a, b, c and d that would match the conditions we set in eqs. (7/12) and (7/13). In order to obtain these coefficients, Blasius simply calibrated his model using known boundary conditions at the edges of the boundary layer: for y = 0 (meaning Y = 0), we have both u = 0 and v = 0. Equation 7/12 becomes: 0 = U du dx + µ ρ д (0) (7/16) 153

154 for y = δ (meaning Y = 1) we have u = U, u/ y = 0 and 2 u/( y) 2 = 0. These values allow us to evaluate three properties of function д : u = U = Uд (1) (7/17) u y = 0 = U δ д (1) (7/18) 2 u ( y) 2 = 0 = U δ 2д (1) (7/19) This system of four equations allowed Pohlhausen to evaluate the four terms a, b, c and d. He introduced the variable Λ, a non-dimensional measure of the outer velocity flow-wise gradient: Λ δ 2 ρ µ du dx (7/20) and he then showed that his velocity model in the smooth-surface laminar boundary layer, equation 7/15, had finally become: u U = 1 (1 + Y )(1 Y )3 + Λ Y 6 (1 Y 3 ) (7/21) Pohlhausen s model is remarkably close to the exact solution described by Blasius (fig. 7.6). We thus have an approximate model for u (x,y) within the boundary layer, as a function of U (x). Figure 7.6 Velocity profiles in the Pohlhausen and Blasius boundary layer profiles. Figure from Richecœeur 2012 [17] It can be seen now that neither Blasius nor Pohlhausen s models are fully satisfactory. The first is drawn from a handful of realistic, known boundary conditions, but we fail to find an analytical solution to match them. The second is drawn from the hypothesis that the analytical solution is simple, but it is not very accurate. And we have not even considered yet the case for turbulent flow! Here, it is the process which matters to us. Both of those methods are frequently-attempted, sensible approaches to solving problems in modern fluid dynamics research, where, just as here, one often has to contend with approximate solutions. 154

155 7.4 Transition After it has traveled a certain length along the wall, the boundary layer becomes very unstable and it transits rapidly from a laminar to a turbulent regime (fig. 7.7). We have already briefly described the characteristics of a turbulent flow in chapter 5; the same apply to turbulence within the boundary layer. Again, we stress that the boundary layer may be turbulent in a globally laminar flow (it may conversely be laminar in a globally turbulent flow): that is commonly the case around aircraft in flight, for example. Figure 7.7 Transition of a boundary layer from laminar to turbulent regime. Figure CC-by Olivier Cleynen We observe that the distance x transition at which the boundary layer changes regime is reduced when the velocity is increased, or when the viscosity is decreased. In practice this distance depends on the distance-based Reynolds number [Re] x ρu x µ. The most commonly accepted prediction for the transition position is: [Re] x transition (7/22) Transition can be generated earlier if the surface roughness is increased, or if obstacles (e.g. turbulators, vortex generators, trip wires) are positioned within the boundary layer. Conversely, a very smooth surface and a very steady, uniform incoming flow will result in delayed transition. 7.5 The turbulent boundary layer The extensive description of turbulent flows remains an unsolved problem. As we have seen in chapter 5 when studying duct flows, by contrast with laminar counterparts turbulent flows result in increased mass, energy and momentum exchange; increased losses to friction; apparently chaotic internal movements. Instead of resolving the entire time-dependent flow in the boundary layer, we satisfy ourselves with describing the average component of the longitudinal speed, u. A widely-accepted velocity model is: u ( y ) 1 U 7 (7/23) δ 155

156 for turbulent boundary layer flow over a smooth surface. This profile has a much flatter geometry near the wall than its laminar counterpart (fig. 7.8). In the same way that we have worked with the laminar boundary layer profiles, we can derive models for our characteristics of interest from this velocity profile: δ x 0,16 7 [Re] 1 x δ x 0,02 7 [Re] 1 x θ x 0,016 7 [Re] 1 x c f(x ) 0,027 [Re] 1 7 x (7/23a) (7/23b) (7/23c) (7/23d) 156 Figure 7.8 Comparison of laminar and turbulent boundary layer profiles, both scaled to the same size. It is important to remember that in practice turbulent boundary layers are much thicker than laminar ones, and that u is only a timeaverage velocity. Figure CC-0 o.c.

157 7.6 Separation Principle Under certain conditions, fluid flow separates from the wall. The boundary layer then disintegrates and we observe the appearance of a turbulent wake near the wall. Separation is often an undesirable phenomenon in fluid mechanics: it may be thought of as the point where we fail to impart a desired trajectory to the fluid. When the main flow speed U along the wall is varied, we observe that the geometry of the boundary layer changes. The greater the longitudinal speed gradient ( du / dx > 0), and the flatter the profile becomes. Conversely, when the longitudinal speed gradient is negative, the boundary layer velocity profile straightens up. When it becomes perfectly vertical at the wall, it is such that streamlines separate from the wall: this is called separation (fig. 7.9). The occurrence of separation can be predicted if we have a robust model for the velocity profile inside the boundary layer. For this, we go back to fundamentals, stating that at the separation point, the shear effort on the surface must be zero: ( ) u τ wall at separation = 0 = µ (7/24) y y=0 At the wall surface (u = 0 and v = 0), equation 7/12 p.152 becomes: ( 2 ) u µ ( y) 2 = dp du = ρu dx dx y=0 (7/25) Thus, as we progressively increase the term dp/ dx, the term 2 u ( y) 2 reaches higher (positive) values on the wall surface. Nevertheless, we know that it must take a negative value at the exterior boundary of the boundary layer. Figure 7.9 Separation of the boundary layer. The main flow is from left to right, and flowing towards a region of increasing pressure. For clarity, the y-scale is greatly exaggerated. Figure CC-by Olivier Cleynen 157

158 Therefore, it must change sign somewhere in the boundary. This point where 2 u changes sign is called inflexion point. ( y) 2 The existence of the inflection point within the boundary layer tells us that the term u y tends towards a smaller and smaller value at the wall. Given enough distance x, it will reach zero value, and the boundary layer will separate (fig. 7.10). Therefore, the longitudinal pressure gradient, which in practice determines the longitudinal velocity gradient, is the key factor in the analytical prediction of separation. We shall remember two crucial points regarding the separation of boundary layers: Video: a football with a rough edge on one side will see turbulent boundary layer on that side and laminar boundary layer on the other. When the laminar layer separates before the turbulent one, the main flow around the ball becomes asymmetric and deviates the ball sideways, a phenomenon that can be exploited to perform impressive tricks. by freekikerz (styl) 1. Separation occurs in the presence of a positive pressure gradient, which is sometimes named adverse pressure gradient. Separation points along a wall (e.g. a car bodywork, an aircraft wing, rooftops, mountains) are always situated in regions where pressure increases (positive dp/ dx or negative du / dx). If pressure remains constant, or if it decreases, then the boundary layer cannot separate. 2. Laminar boundary layers are much more sensitive to separation than turbulent boundary layers (fig. 7.11). A widely-used technique to reduce or delay the occurrence of separation is to make boundary layers turbulent, using low-height artificial obstacles positioned in the flow. By doing so, we increase shear-based friction (which increases with turbulence) as a trade-off for better resistance to stall. Predicting in practice the position of a separation point is difficult, because an intimate knowledge of the boundary layer profile and of the (externalflow-generated) pressure field are required and as the flow separates, these are no longer independent. Resorting to experimental measurements, in this case, is often a wise idea! 158 Figure 7.10 The inflection point within a boundary layer about to separate. Figure CC-by Olivier Cleynen

159 Figure 7.11 The effect of decreasing Reynolds number on flow attachment over an airfoil at constant angle of attack, with the transition point highlighted. Laminar boundary layers are much more prone to separation than turbulent boundary layers. Figure CC-by-sa Olivier Cleynen, based on Barlow & Pope 1999 [9] Separation according to Pohlhausen In the case of a laminar flow along a smooth surface, Pohlhausen s model allows for interesting results. Indeed, at the separation point, τ wall = 0, thus: ( ) u ( U ) τ wall = 0 = µ = µ y δ д y=0 0 = д Y =0 = a Y =0 Λ = 12 (7/26) δ 2 ρ µ du dx = 12 (7/27) This equation allows us to predict the thickness of the boundary layer at the point where separation occurs, and therefore its position, if the velocity gradient du / dx is known. We can see here that once we have obtained an extensive description of the flow within the boundary layer, we can generate a large amount of useful information regarding the general flow around the object. Unfortunately, Pohlhausen s model s agreement with experimental measurements is not very good, and it can only usefully predict the position of the separation point in a few simple cases. Nevertheless, the overall thought process, i.e. intuition of a simple model non-dimensionalization correlation with known boundary conditions 159

160 exploitation of the model to predict flow behavior is what should be remembered here, for it is typical of the research process in fluid dynamics. 160

161 Exercise sheet 7 (Flow near walls) last edited April 4, 2017 These lecture notes are based on textbooks by White [13], Çengel & al.[16], and Munson & al.[18]. Except otherwise indicated, we assume that fluids are Newtonian, and that: ρ water = kg m 3 ; p atm. = 1 bar; ρ atm. = 1,225 kg m 3 ; T atm. = 11,3 C; µ atm. = 1, N s m 2 ; д = 9,81 m s 2. Air is modeled as a perfect gas (R air = 287 J K 1 kg 1 ; γ air = 1,4; c pair = J kg 1 K 1 ). In boundary layer flow, we accept that transition occurs at [Re] x The wall shear coefficient c f, a function of distance x, is defined based on the free-stream flow velocity U : c f(x ) τ wall 1 2 ρu 2 (7/5) Exact solutions to the laminar boundary layer along a smooth surface yield: δ x = 4,91 δ [Re]x x = 1,72 [Re]x θ x = 0,664 c f(x ) = 0,664 [Re]x [Re]x (7/14b) (7/14d) Solutions to the turbulent boundary layer along a smooth surface yield the following time-averaged characteristics: δ x 0,16 [Re] 1 7 x δ x 0,02 [Re] 1 7 x θ x 0,016 [Re] 1 7 x c f(x ) 0,027 [Re] 1 7 x (7/23b) (7/23d) Figure 7.12 quantifies the viscosity of various fluids as a function of temperature. 161

162 Figure 7.12 Viscosity of various fluids at a pressure of 1 bar (in practice viscosity is almost independent of pressure). Figure repeated from fig p.119; White 2008 [13] 7.1 Water and air flow White [13] E7.2 A flat plate of length 0,3 m is placed parallel to a uniform flow with speed 0,3 m s 1. How thick can the boundary layer become: 1. if the fluid is air at 1 bar and 20 C? 2. if the fluid is water at 20 C? 7.2 Boundary layer sketches CC-0 o.c. A thin and long horizontal plate is moved horizontally through a stationary fluid. 1. Sketch the velocity profile of the fluid: at the leading edge; at a point where the boundary layer is laminar; and at a point further downstream where the boundary layer is turbulent. 2. Draw a few streamlines, indicate the boundary layer thickness δ, and the displacement thickness δ. 3. Explain shortly (e.g. in 30 words or less) how the transition to turbulent regime can be triggered. 4. Explain shortly (e.g. in 30 words or less) how the transition to turbulent regime could instead be delayed. 162

163 7.3 Shear force due to boundary layer White [13] E7.3 A thin and smooth plate of dimensions 0,5 3 m is placed with a zero angle of attack in a flow incoming at 1,25 m s 1, as shown in fig Figure 7.13 A thin plate positioned parallel to an incoming uniform flow. Two configurations are studied in this exercise. Figure CC-0 o.c. 1. What is the shear force exerted on the plate for each of the two configurations shown in fig. 7.13, when the fluid is air of viscosity µ air = 1, Pa s? 2. What are the shear forces when the fluid is water of viscosity µ water = Pa s? 3. [difficult question] How would these shear efforts evolve if the plate was tilted with an angle of 20 relative to the flow? 7.4 Wright Flyer I CC-0 o.c. The Wright Flyer I, the first airplane capable of sustained controlled flight (1903), was a biplane with a 12 m wingspan (fig. 7.14). It had two wings of chord length 1,98 m stacked one on top of the other. The wing profile was extremely thin and it could only fly at very low angles of attack. Its flight speed was approximately 40 km h 1. Figure 7.14 The Wright Flyer I, first modern airplane in history. Built with meticulous care and impeccable engineering methodology by two bicycle makers, it made history in December Photo by Orville Wright, 1908 (public domain) 1. If the flow over the wings can be treated as if they were flat plates, what is the power necessary to compensate the shear exerted by the airflow on the wings during flight? 2. Which other forms of drag would also be found on the aircraft? (give a brief answer, e.g. in 30 words or less) 163

164 7.5 Power lost to shear on an airliner fuselage CC-0 o.c. An Airbus A is cruising at [Ma] = 0,82 at an altitude of m (where the air has viscosity 1, N s m 2, temperature 220 K, density 0,4 kg m 3 ). The cylindrical part of the fuselage has diameter 5,6 m and length 65 m. 1. What is approximately the maximum boundary layer thickness around the fuselage? 2. What is approximately the average shear applying on the fuselage skin? 3. Estimate the power dissipated to friction on the cylindrical part of the fuselage. 4. In practice, in which circumstances could flow separation occur on the fuselage skin? (give a brief answer, e.g. in 30 words or less) 7.6 Separation according to Pohlhausen non-examinable, based on Richecœur 2012 [17] Air at 1 bar and 20 C flows along a smooth surface, and decelerates slowly, with a constant rate of 0,25 m s 1 m 1. According to the Pohlhausen model (eq. 7/21 p.154), at which distance downstream will separation occur? Is the boundary layer still laminar then? How could one generate such a deceleration in practice? 7.7 Laminar wing profile non-examinable. Based on a diagram from Bertin et al [15] The characteristics of a so-called laminar wing profile are compared in figs to 7.17 with those of an ordinary profile. On the graph representing the pressure coefficient C p p p 1 2 ρv 2, identify the curve corresponding to each profile. What advantages and disadvantages does the laminar wing profile have, and how can they be explained? In which applications will it be most useful? 164 Figure 7.15 Comparison of the thickness distribution of two uncambered wing profiles: an ordinary medium-speed naca 0009 profile, and a laminar naca profile. Figure Bertin & Cummings 2010 [15]

165 Figure 7.16 Static pressure distribution (represented as a the local non-dimensional pressure coefficient C p p p 1 2 ρv ) as a function of distance x (non-dimensionalized with the chord c) over 2 the surface of the two airfoils shown in fig Figure Bertin & Cummings 2010 [15] Figure 7.17 Values of the section drag coefficient C d 1 2 c ρv as a function of the section lift 2 l coefficient C l 1 2 c ρv for both airfoils presented in fig d Figure Bertin & Cummings 2010 [15], based on data by Abott & Von Doenhoff 1949 [1] 165

166 7.8 Separation mechanism non-examinable, CC-0 o.c. Sketch the velocity profile of a laminar or turbulent boundary layer shortly upstream of, and at a separation point. The two equations below describe flow in laminar boundary layer: u u x + v u y = U du 2 u ( y) 2 (7/12) dx + µ ρ u x + v y = 0 (7/13) Identify these two equations, list the conditions in which they apply, and explain shortly (e.g. in 30 words or less) why a boundary layer cannot separate when a favorable pressure gradient is applied along the wall. 7.9 Thwaites separation model non-examinable, from White [13] E7.5 In 1949, Bryan Thwaites explored the limits of the Pohlhausen separation model. He proposed a different model to describe the laminar boundary layer velocity profile, which is articulated upon an expression for the momentum thickness θ and is more accurate than the Reynolds-number based descriptions that we have studied. Thwaites proposed the model: ( θ 2 = θ0 2 U0 ) 6 + 0,45 µ 1 x U ρ U 6 U 5 dx (7/28) in which the 0-subscripts denote initial conditions (usually θ 0 = 0) Thwaites also defined a non-dimensional momentum thickness λ θ : 0 λ θ θ 2 ρ µ du dx (7/29) and then went on to show that for the model above, separation (c f = 0) occurs when λ θ = 0,09. A classical ( model ) for flow deceleration is the Howarth longitudinal velocity profile U (x) = U 0 1 x L, in which L is a reference length of choice. In this velocity distribution with linear deceleration, what is the distance at which the Thwaites model predicts the boundary layer separation? 166

167 Answers 7.1 1) At trailing edge [Re] x = thus the layer is laminar everywhere. δ will grow from 0 to 2,01 cm (eq. 7/14a p.153); 2) For water: δ trailing edge = 4,91 mm ) See fig. 7.7 p.155. At the leading-edge the velocity is uniform. Note that the y-direction is greatly exaggerated, and that the outer velocity U is identical for both regimes; 2) See fig. 7.4 p.150. Note that streamlines penetrate the boundary layer; 3) and 4) See 7.4 p x transition, air = 4,898 m and x transition, water = 0,4 m. In a laminar boundary layer, inserting equation 7/14d into equation 7/5 into equation 7/6 yields F τ = 0,664LU 1,5 [ ] xtransition ρµ x. 0 In a turbulent boundary [ layer, we use equation 7/23d instead and get F τ = 0,01575Lρ 7U µ 1 6 ] xtrailing edge 7 x 7. These expressions allow the calculation of x transition the forces below, for the top surface of the plate: 1) (air) First case: F = 3, N; second case F = 8, N (who would have thought eh?); 2) (water) First case: F = 3,7849 N; second case F = 2,7156 N. 7.4 Using the expressions developed in exercise 7.3, W friction 255 W ) x transition = 7,47 cm (the laminar part is negligible). With the equations developed in exercise 7.3, we get F = 24,979 kn and W = 6,09 MW. Quite a jump from the Wright Flyer I! 2) When the longitudinal pressure gradient is zero, the boundary layer cannot separate. Thus separation from the fuselage skin can only happen if the fuselage is flown at an angle relative to the flight direction (e.g. during a low-speed maneuver). 7.9 Once the puzzle pieces are put together, this is an algebra exercise: ( ) x L = separation 0,1231. Bryan beats Ernst! 167

168 168

169 Fluid Mechanics Chapter 8 Modeling large- and small-scale flows last edited October 30, Motivation Flow at large scales Problem statement Investigation of inviscid flows Plotting velocity with functions Kinematics that fit conservation laws Strengths and weaknesses of potential flow Superposition: the lifting cylinder Circulating cylinder 175 Circulating cylinder Modeling lift with circulation Flow at very small scales Exercises 183 These lecture notes are based on textbooks by White [13], Çengel & al.[16], and Munson & al.[18]. 8.1 Motivation This exploratory chapter is not a critical component of fluid dynamics; instead, it is meant as a brief overview of two extreme cases: flows for which viscous effects are negligible, and flows for which they are dominant. This exploration should allow us to answer two questions: How can we model large-scale flows analytically? How can we model small-scale flows analytically? Video: pre-lecture briefing for this chapter by o.c. (CC-by) Flow at large scales Problem statement In this section, we are interested in flow at very large scales: those for which a representative length L is very large. In particular, when L is large enough, the influence of viscosity is reduced. Formally, this corresponds to the case where the Reynolds number [Re] ρv L/µ is very large. To examine the mechanics of such a flow, we turn to our beloved nondimensional Navier-Stokes equation for incompressible flow derived as eq. 6/17 p.135, [St] V t + [1] V V = 1 [Fr] 2 д [Eu] p + 1 [Re] 2 V (8/1) 169

170 in which we see that when the Reynolds number [Re] is very large, the last term becomes negligible relative to the other four terms. Thus, our governing equation can be reduced as follows: [St] V t + [1] V V 1 [Fr] 2 д [Eu] p (8/2) Now, converting eq. 8/2 back to dimensional terms, our governing momentum equation for inviscid flow becomes: ρ D V Dt = ρ д p (8/3) We see that with the starting proposition that [Re] was large, we have removed altogether the viscous (last) term from the Navier-Stokes equation. Flows governed by this equation are called inviscid flows. Equation 8/3 is named the Euler equation; it stipulates that the acceleration field is driven only by gravity and by the pressure field Investigation of inviscid flows From what we studied in chapter 6, we recognize immediately that flows governed by eq. 8/3 are troublesome: the absence of viscous effects facilitates the occurrence of turbulence and makes for much more chaotic behaviors. Although the removal of shear from the Navier-Stokes equation simplifies the governing equation, the solutions to this new equation become even harder to find and describe. What can be done in the other two branches of fluid mechanics? Large-scale flows are difficult to investigate experimentally. As we have seen in chapter 6, scaling down a flow (e.g. so it may fit inside a laboratory) while maintaining constant [Re] requires increasing velocity by a corresponding factor. Large scale flows are also difficult to investigate numerically. At high [Re], the occurrence of turbulence makes for either an exponential increase in computing power (direct numerical simulation computing power increases with [Re] 3,5 ) or for increased reliance on hard-tocalibrate turbulence models (in Reynolds-averaged simulations). All three branches of fluid mechanics, therefore, struggle with turbulent flows. In spite of this, large-scale flows are undeniably important. In chapter 7 we were able to understand and describe fluid flow very close to walls. Now we wish to be able to to the same for large structures, for example, in order to describe the broad patterns of fluid flow (in particular, pressure distribution) in the wake of an aircraft or of a wind turbine, or within a hurricane. It is clear that we have no hope of accounting easily for turbulence, but we can at least describe the main features of such flows by restricting ourselves to laminar cases. In the following sections, we will model such laminar solutions directly, based on intuition and observation, and make sure that they match the condition described by eq. 8/3 above. 170

171 8.3 Plotting velocity with functions Kinematics that fit conservation laws For simple flow structures, the velocity field can be simply described based on observation and intuition. This is done for example in exercise 8.8, where we reconstruct the flow field within and around a tornado using very simple, almost primitive, kinematics. Without so much as a small increase in complexity, this approach becomes untenable. In the last exercises of chapter 4 (ex. 4.7 & 4.8 p.105) we have seen that it is easy to propose a velocity field that does not satisfy (is not a solution of) either the continuity or the Navier-Stokes equation. For example, if one considers two of the tornado flows mentioned above together, the velocity field cannot be described easily anymore. One approach has been developed in the 17 th century to overcome this problem. It consists in finding a family of flows, all steady, that always satisfy the conservation equations. Those flows can then be added to one another to produce new conservation-abiding flows. Such flows are called potential flows. Two conditions need to be fulfilled for this approach to work: 1. The velocity field must always be describable with a function; that is, there must correspond a single value of u, of v and of w at each of the coordinates x i, y i, or z i in space. This means in practice that we cannot account for flows which curl up on themselves, occasionally recirculating back on their path. 2. The velocity field must conserve mass. In incompressible flow, this is achieved if the continuity equation V = 0 (eq. 4/17 p.90) is respected. With potential flow, these two conditions are addressed as follows: 1. We restrict ourselves to irrotational flows, those in which the curl of velocity (see Appendix A2 p.217) is always null: by definition, for an irrotational flow. V = 0 (8/4) It can be shown that flows are irrotational when there exists a scalar function ϕ (pronounced phi and named potential function) of which the gradient is the velocity vector field: ϕ V (8/5) In the case of two-dimensional flow, this translates as: ϕ x u (8/6) ϕ y v (8/7) When plotted out, lines of constant ϕ (named equipotential lines) are always perpendicular to the streamlines of the flow. 171

172 2. The continuity equation is satisfied by referring to stream functions. It can be shown that the divergent of velocity is null when there exists a vector field function ψ (pronounced psi and named stream function) of which the curl is the velocity vector field: ψ = V (8/8) In the case of two-dimensional flow, ψ is a scalar field and eq. 8/8 translates as: ψ y u (8/9) ψ x v (8/10) When plotted out, lines of of constantψ value are streamlines in other words, as they travel along, fluid particles follow paths of constant ψ value. In summary, we have shifted the problem from looking for u and v, to looking for ψ and ϕ. The existence of such functions ensures that flows can be added and subtracted from one another yet will always result in mass-conserving, mathematically-describable flows. If such two functions are known, then the velocity components can be obtained easily either in Cartesian coordinates, u = ϕ x = ψ y v = ϕ y = ψ x (8/11) (8/12) or angular coordinates: v r = ϕ r = 1 r v θ = 1 r ψ θ ϕ θ = ψ r (8/13) (8/14) Strengths and weaknesses of potential flow The potential flow methodology allows us to find solutions to the Euler equation: flows in which the Reynolds number is high enough that viscosity has no significant role anymore. Potential flows are the simplest solutions that we are able to come up with. They are: strictly steady; inviscid; incompressible; devoid of energy transfers; two-dimensional (in the scope of this course at least). 172 This is quite convenient for the academician, who recognizes immediately four of the five criteria which we set forth in chapter 3 for using the Bernoulli

173 equation. Along a streamline, the fifth condition is met, and Euler s equation reduces to eq. 3/18 (p.69): p 1 ρ V дz 1 = p 2 ρ V дz 2 = cst. (8/15) along a streamline in a steady incompressible inviscid flow. and so it follows that if the solution to an inviscid flow is known, the forces due to pressure can be calculated with relative ease. Nevertheless, from a science and engineering point of view, potential flows have only limited value, because they are entirely unable to account for turbulence, which we have seen is an integral feature of high-[re] flows. We should therefore use them only with great caution. Potential flows help us model large-scale structures with very little computational cost, but this comes with strong limitations Superposition: the lifting cylinder It is possible to describe a handful of basic potential flows called elementary flows as fundamental ingredients that can be added to one another to create more complex and interesting flows. Without going into much detail, the most relevant elementary flows are: Uniform longitudinal flow, ϕ = V r cos θ (8/16) ψ = V r sin θ (8/17) Sources and sinks (fig. 8.1) which are associated with the appearance of a (positive or negative) volume flow rate V from a single point in the flow: ϕ = ψ = V ln r 2π (8/18) V 2π θ (8/19) Irrotational vortices, rotational patterns which impart a rotational velocity v θ = f (r,θ ) on the flow, in addition to which there may exist Figure 8.1 Concept of a source inside a two-dimensional potential flow. A sink would display exactly opposed velocities. Figure CC-by-sa Commons User:Nicoguaro 173

174 a radial component v r : ϕ = Γ 2π θ (8/20) ψ = Γ ln r 2π (8/21) in which Γ (termed circulation) is a constant proportional to the strength of the vortex. Doublets, which consist in a source and a sink of equal volume flow rate positioned extremely close one to another (fig. 8.2): ϕ = K cos θ r ψ = K sin θ r (8/22) (8/23) in which K is a constant proportional to the source/sink volume flow rate V. It was found in the 17 th Century that combining a doublet with uniform flow resulted in flow patterns that matched that of the (inviscid, steady, irrotational) flow around a cylinder (fig. 8.3). That flow has the stream function: ) ψ = U sin θ (r R2 (8/24) r This stream function allows us to describe the velocity everywhere: u r = 1 ) ψ r θ = U cos θ (1 R2 r 2 u θ = ψ ) r = U sin θ (1 + R2 r 2 (8/25) (8/26) We can even calculate the lift and drag applying on the cylinder surface. Indeed, along the cylinder wall, r = R and u r r=r = 0 (8/27) u θ r=r = 2U sin θ (8/28) Since the Bernoulli equation can be applied along any streamline in this (steady, constant-energy, inviscid, incompressible) flow, we can express the Figure 8.2 Left: a simple uniform steady flow; Right: a doublet, the result of a source and a sink brought very close one to another Figure CC-0 o.c. 174

175 Figure 8.3 The addition of a doublet and a uniform flow produces streamlines for an (idealized) flow around a cylinder. Figure CC-by-sa by Commons User:Kraaiennest pressure p s on the cylinder surface as a function of θ: p ρu 2 = p s ρv2 θ p s (θ ) = p ρ ( U 2 v 2 θ ) (8/29) Now a simple integration yields the net forces exerted by the fluid on the cylinder per unit width L, in each of the two directions x and y: F net,x L F net,y L = = 2π 0 2π 0 p s cos θ R dθ = 0 (8/30) p s sin θ R dθ = 0 (8/31) The results are interesting, and at the time they were obtained by their author, Jean le Rond D Alembert, were devastating: both lift and drag are zero. This inability to reproduce the well-known phenomena of drag is often called the d Alembert paradox. The pressure distribution on the cylinder surface can be represented graphically and compared to experimental measurements (fig. 8.4). Good agreement is obtained with measurements on the leading edge of the cylinder; but as the pressure gradient becomes unfavorable, in practice the boundary layer separates a phenomenon that cannot be described with inviscid flow and a low-pressure area forms on the downstream side of the cylinder. 175

176 Figure 8.4 Pressure distribution (relative to the far-flow pressure) on the surface of a cylinder, with flow from left to right. On the left is the potential flow case, purely symmetrical. On the right (in blue) is a measurement made at a high Reynolds number. Boundary layer separation occurs on the second half of the cylinder, which prevents the recovery of leading-edge pressure values, and increases drag. Figure CC-by-sa Commons User:BoH & Olivier Cleynen Circulating cylinder Video: watch the Brazilian football team show their French counterparts how circulation (induced through friction by ball rotation) is associated to dynamic lift on a circular body by TF1, 1997 (styl) An extremely interesting hack can be played on the potential cylinder flow above if an irrotational vortex of stream function ψ = Γ 2π ln r (eq. 8/21) is added to it. The overall flow field becomes: ) ψ = U sin θ (r R2 Γ ln r (8/32) r 2π With this function, several key characteristics of the flow field can be obtained. The first is the velocity field at the cylinder surface: u r r=r = 0 (8/33) u θ r=r = 2U sin θ + Γ 2πR (8/34) and we immediately notice that the velocity distribution is no longer symmetrical with respect to the horizontal axis (fig. 8.5): the fluid is deflected, and so there will be a net force on the cylinder. This time, the net pressure forces on the cylinder have changed: F net,x L F net,y L = = 2π 0 2π 0 p s cos θ R dθ = 0 (8/35) p s sin θ R dθ = ρ U Γ (8/36) 176 We thus find out that the drag is once again zero as for any potential flow but that lift occurs which is proportional to the free-stream velocity U and to the circulation Γ. In practice, such a flow can be generated by spinning a cylindrical part in a uniform flow. A lateral force is then obtained, which can be used as a propulsive or sustaining force. Several boats and even an aircraft have been used in practice to demonstrate this principle. Naturally, flow separation from the cylinder profile and the high shear efforts generated on the surface

177 Figure 8.5 The addition of an irrotational vortex on top of the cylinder flow described in fig. 8.3 distorts the flow field and it becomes asymmetrical: a lift force is developed, which depends directly on the circulation Γ. Figure White 2008 [13] cause real flows to differ from the ideal case described here, and it turns out that rotating cylinders are a horribly uneconomical and unpractical way of generating lift Modeling lift with circulation Fluid flow around cylinders may have little appeal for the modern student of fluid mechanics, but the methodology above has been taken much further. With further mathematical manipulation called conformal mapping, potential flow can be used to described flow around geometrical shapes such as airfoils (fig. 8.6). Because the flow around such streamlined bodies usually does not feature boundary layer separation, the predicted flow fields everywhere except in the close vicinity of the solid surface are accurately predicted. There again, it is observed that regardless of the constructed geometry, no lift can be modeled unless circulation is also added within the flow. The amount of circulation needed so that results may correspond to experimental observations is found by increasing it progressively until the the rear stagnation point reaches the rear trailing edge of the airfoil, a condition known as the Kutta condition. Regardless of the amount of circulation added, potential flow remains entirely reversible, both in a kinematic and a thermodynamic sense, thus, care must be taken in the problem setup to make sure the model is realistic (fig. 8.7). It is then observed in general that any dynamic lift generation can be modeled as the superposition of a free-stream flow and a circulation effect (fig. 8.8). With such a tool, potential flow becomes an extremely useful tool, mathematically and computationally inexpensive, in order to model and understand the cause and effect of dynamic lift in fluid mechanics. In particular, it has been paramount in the description of aerodynamic lift distribution over aircraft wing surfaces (fig. 8.9), with a concept called the Lifting-line theory. Video: acting on swimming pool water with a round plate sheds a half-circular vortex that is extremely stable and can be interacted with quite easily. Such stable laminar structures are excellent candidates for analysis using potential flow. by Physics Girl (Dianna Cowern) (styl) 177

178 Figure 8.6 Potential flow around an airfoil without (top) and with (bottom) circulation. Much like potential flow around a cylinder, potential flow around an airfoil can only result in a net vertical force if an irrotational vortex (with circulation Γ) is added on top of the flow. Only one value for Γ will generate a realistic flow, with the rear stagnation point coinciding with the trailing edge, a occurrence named Kutta condition. Figure CC-0 o.c. Figure 8.7 Potential flow allows all velocities to be inverted without any change in the flow geometry. Here the flow around an airfoil is reversed, displaying unphysical behavior. Figure CC-0 o.c. Figure 8.8 A numerical model of flow around an airfoil. In the left figure, the velocity vectors are represented relative to a stationary background. In the right figure, the velocity of the free-stream flow has been subtracted from each vector, bringing the circulation phenomenon into evidence. Figures 1 & 2 CC-by-sa by en:wikipedia User:Crowsnest 178

179 Figure 8.9 From top to bottom, the lift distribution over the wings of a glider is modeled with increasingly complex (and accurate) lift and circulation distributions along the span. The Lifting-line theory is a method associating each element of lift with a certain amount of circulation. The effect of each span-wise change of circulation is then mapped onto the flow field as a trailing vortex. Figures 1, 2 & 3 CC-by-sa Olivier Cleynen 179

180 8.4 Flow at very small scales At the complete opposite of the spectrum, we find flow at very small scales: flows around bacteria, dust particles, and inside very small ducts. In those flows the representative length L is extremely small, which makes for small values of the Reynolds number. Such flows are termed creeping or Stokes flows. What are their main characteristics? Looking back once again at the non-dimensional Navier-Stokes equation for incompressible flow derived as eq. 6/17 p.135, [St] V t + [1] V V = 1 [Fr] 2 д [Eu] p + 1 [Re] 2 V (8/37) we see that creeping flow will occur when the Reynolds number is much smaller than 1. The relative weight of the term 1 [Re] 2 V then becomes overwhelming. In addition to cases where [Re] 1, we focus our interest on flows for which: gravitational effects have negligible influence over the velocity field; the characteristic frequency is extremely low (quasi-steady flow). With these characteristics, the terms associated with the Strouhal and Froude numbers become very small with respect to the other terms, and our nondimensionalized Navier-Stokes equation (eq. 8/37) is approximately reduced to: 0 [Eu] p + 1 [Re] 2 V (8/38) We can now come back to dimensionalized equations, concluding that for a fluid flow dominated by viscosity, the pressure and velocity fields are linked together by the approximate relation: p = µ 2 V (8/39) In this type of flow, the pressure field is entirely dictated by the Laplacian of velocity, and the fluid density has no importance. Micro-organisms, for which the representative length L is very small, spend their lives in such flows (fig. 8.10). At the human scale, we can visualize the effects of these flows by moving an object slowly in highly-viscous fluids (e.g. a spoon in honey), or by swimming in a pool filled with plastic balls. The inertial effects are almost inexistent, drag is extremely important, and the object geometry has comparatively small influence. In 1851, George Gabriel Stokes worked through equation 8/39 for flow around a sphere, and obtained an analytical solution for the flow field. This allowed him to show that the drag F D sphere applying on a sphere of diameter D in creeping flow (fig. 8.11) is: F D sphere = 3π µu D (8/40) 180

181 Figure 8.10 Micro-organisms carry themselves through fluids at extremely low Reynolds numbers, since their scale L is very small. For them, viscosity effects dominate inertial effects. Photo by Yutaka Tsutsumi, M.D., Fujita Health University School of Medicine Inserting this equation 8/40 into the definition of the drag coefficient C FD F D / 1 2 ρs frontalu 2 (from eq. 6/2 p.130) then yields: C FD = F D sphere 24µ 1 2 ρu 2 π = 4 D2 ρ U D = 24 (8/41) [Re] D These equations are specific to flow around spheres, but the trends they describe apply well to most bodies evolving in highly-viscous flows, such as dust or liquid particles traveling through the atmosphere. Drag is only proportional to the speed (as opposed to low-viscosity flows in which it grows with velocity squared), and it does not depend on fluid density. Figure 8.11 Flow at very low Reynolds numbers around a sphere. In this regime, the drag force is proportional to the velocity. Figure CC-by-sa by Olivier Cleynen & Commons User:Kraaiennest 181

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183 Exercise sheet 8 (Modeling large- and small-scale flows) last edited June 18, 2017 These lecture notes are based on textbooks by White [13], Çengel & al.[16], and Munson & al.[18]. Except otherwise indicated, we assume that fluids are Newtonian, and that: ρ water = kg m 3 ; p atm. = 1 bar; ρ atm. = 1,225 kg m 3 ; T atm. = 11,3 C; µ atm. = 1, N s m 2 ; д = 9,81 m s 2. Air is modeled as a perfect gas (R air = 287 J K 1 kg 1 ; γ air = 1,4; c pair = J kg 1 K 1 ). In a highly-viscous (creeping) steady flow, the drag F D exerted on a spherical body of diameter D at by flow at velocity U is quantified as: F D sphere = 3π µu D (8/40) 8.1 Volcanic ash from the Eyjafjallajökull Çengel & al. [16] E10.2 In 2010, a volcano with a complicated name and unpredictable mood decided to ground the entire European airline industry for five days. We consider a microscopic ash particle released at very high altitude ( 50 C, 0,55 bar, 1, N s m 2 ). We model it as a sphere with 50 µm diameter. The density of volcanic ash is kg m What is the terminal velocity of the particle? 2. Will this terminal velocity increase or decrease as the particle progresses towards the ground? (give a brief, qualitative answer) 8.2 Water drop Çengel & al. [16] A rainy day provides yet another opportunity for exploring fluid dynamics (fig. 8.12). A water drop with diameter 42,4 µm is falling through air at 25 C and 1 bar. 1. Which terminal velocity will it reach? 2. Which velocity will it reach once its diameter will have doubled? 183

184 Figure 8.12 A sketched diagram showing the geometry of water drops of various sizes in free fall. When their diameter is lower than 2 mm, water drops are approximately spherical (B). As they grow beyond this size, their shape changes and they eventually break-up (C-E). They never display the classical shape displayed in A, which is caused only by surface tension effects when they drip from solid surfaces. Figure CC-by-sa by Ryan Wilson 8.3 Hangar roof based on White [13] P8.54 Certain flows in which both compressibility and viscosity effects are negligible can be described using the potential flow assumption (the hypothesis that the flow is everywhere irrotational). If we compute the two-dimensional laminar steady fluid flow around a cylinder profile, we obtain the velocities in polar coordinates as: u r = 1 r ψ θ = U cos θ u θ = ψ r = U sin θ ) (1 R2 r 2 ) (1 + R2 where the origin (r = 0) is at the center of the cylinder profile; θ is measured relative to the free-stream velocity vector; U is the incoming free-stream velocity; and R is the (fixed) cylinder radius. Based on this model, in this exercise, we study the flow over a hangar roof. r 2 (8/25) (8/26) Wind with a nearly-uniform velocity U = 100 km h 1 is blowing across a 50 m-long hangar with a semi-cylindrical geometry, as shown in fig The radius of the hangar is R = 20 m. 1. If the pressure inside the hangar is the same as the pressure of the faraway atmosphere, and if the wind closely follows the hangar roof geometry (without any flow separation), what is the total lift force on the hangar? (hint: we accept that sin 3 x dx = 1 3 cos3 x cos x + k). 184

185 Figure 8.13 A semi-cylindrical hangar roof. Wind with uniform velocity U flows perpendicular to the cylinder axis. Figure CC-0 o.c. 2. At which position on the roof could we drill a hole to negate the aerodynamic lift force? 3. Propose two reasons why the aerodynamic force measured in practice on the hangar roof may be lower than calculated with this model. 8.4 Cabling of the Wright Flyer derived from Munson & al. [18] The Wright Flyer I, the first powered and controlled aircraft in history, was subjected to multiple types of drag. We have already studied viscous friction on its thin wings in exercise 7.4. The data in figure 8.14 provides the opportunity to quantify drag due to pressure. A network of metal cables with diameter 1,27 mm criss-crossed the aircraft in order to provide structural rigidity. The cables were positioned perpendicularly to the air flow, which came at 40 km h 1. The total cable length was approximately 60 m. What was the drag generated by the cables? 185

186 Figure 8.14 Experimental measurements of the drag coefficient applying to a cylinder and to a sphere as a function of the diameter-based Reynolds number [Re] D, shown together with schematic depictions of the flow around the cylinder. By convention, the drag coefficient C D C F D F D 1 2 ρsu 2 (eq. 6/2 p.130) compares the drag force F D with the frontal area S. Both figures from Munson & al.[18] 186

187 8.5 Ping pong ball Munson & al. [18] E9.16 A series of experiments is conducted in a wind tunnel on a large cast iron ball with a smooth surface; the results are shown in fig These measurement data are used to predict the behavior of a ping pong ball. Table tennis regulations constrain the mass of the ball to 2,7 g and its diameter to 40 mm. 1. Is it possible for a ball thrown at a speed of 50 km h 1 to have a perfectly horizontal trajectory? 2. If so, what would be its deceleration? 3. How would the drag and lift applying on the ball evolve if the air viscosity was progressively decreased to zero? Figure 8.15 Experimental measurements of the lift and drag coefficients applying on a rotating sphere in an steady uniform flow. Figure from Munson & al.[18] 187

188 8.6 Lift on a symmetrical object non-examinable Briefly explain (e.g. with answers 30 words or less) how lift can be generated on a sphere or a cylinder, with differential control boundary layer control; with the effect of rotation. Draw a few streamlines in a two-dimensional sketch of the phenomenon. 8.7 Air flow over a wing profile Non-examinable. From Munson & al. [18] The characteristics of a thin, flat-bottomed airfoil are examined by a group of students in a wind tunnel. The first investigations focus on the boundary layer, and the research group evaluate the boundary layer thickness and make sure that it is fully attached. Once this is done, the group proceeds with speed measurements all around the airfoil. Measurements of the longitudinal speed u just above the boundary layer on the top surface are tabulated below: x/c (%) y/c (%) u/u ,5 3,72 0, ,3 1,232 7,5 6,48 1, ,43 1, ,92 1, ,14 1, ,49 1, ,45 1, ,11 1, ,46 1, ,62 0, ,26 0, ,807 On the bottom surface, the speed is measured as being constant (u = U ) to within experimental error. What is the lift coefficient on the airfoil? 188

189 8.8 Flow field of a tornado From Çengel & al. [16] E9-5, E9-14 & E10-3 The continuity and Navier-Stokes equations for incompressible flow written in cylindrical coordinates are as follows: ρ v r t ρ ρ v r + v r r + v θ r = ρд r p r + µ [ vθ t v r θ v2 θ r + v z ( r v ) r r r [ 1 r v θ + v r r + v θ r = ρд θ 1 r [ vz t p θ + µ v z + v r r + v θ r = ρд z p z + µ 1 ru r r r + 1 r v r z v θ θ + v ] rv θ v θ + v z r z [ ( 1 r v ) θ r r r ] v r θ + v v z z z ( r v ) z r r [ 1 r v r r r 2 2 v r ( θ ) 2 2 r 2 v θ θ + 2 v r ( z) 2 v θ r r 2 2 v θ ( θ ) r 2 v r θ + 2 v θ ( z) ] v z r 2 ( θ ) v z ( z) 2 ] ] (8/42) (8/43) (8/44) u θ θ + u z z = 0 (8/45) In this exercise, we are interested in solving the pressure field in a simplified description of a tornado. For this, we consider only a horizontal layer of the flow, and we consider that all properties are independent of the altitude z and of the time t. We start by modeling the tornado as a vortex imparting an angular velocity such that: u θ = Γ 2πr in which Γ is the circulation (measured in s 1 ) and remains constant everywhere. (8/46) 1. What form must the radial velocity u r take in order to satisfy continuity? Among all the possibilities for u r, we choose the simplest in our study, so that: u r = 0 (8/47) all throughout the tornado flow field. 2. What is the pressure field throughout the horizontal layer of the tornado? The flow field described above becomes unphysical in the very center of the tornado vortex. Indeed, our model for u θ is typical of an irrotational vortex, which, like all irrotational flows, is constructed under the premise that the flow is inviscid. However, in the center of the vortex, we are confronted with high velocity gradients over very small distances: viscous effects can no longer be neglected and our model breaks down. It is observed that in most such vortices, a vortex core region forms that rotates just like a solid cylindrical body. This flow is rotational and its governing equations result in a realistic pressure distribution. 189

190 3. What is the pressure field within the rotational core of the tornado? Make a simple sketch showing the pressure distribution as a function of radius throughout the entire tornado flow field. 4. [An opening question for curious students] What is the basic mechanism of vortex stretching? How would it modify the flow field described here? 190

191 Answers 8.1 1) At terminal velocity, the weight of the sphere equals the drag. This allows us to D obtain U = дρ 2 sphere 18µ = 0,1146 m s 1 : unbearably slow when you are stuck in an airport! With U, check that the Reynolds number indeed corresponds to creeping flow: [Re] D = 0, Same as previous exercise: U 1 = 4, m s 1 and U 2 = 0,183 m s 1, with Reynolds numbers of 0,113 and 0,906 respectively (thus creeping flow hypothesis valid) ) express roof pressure as a function of θ using eq. 8/29 p.175 on eq. 8/26, then integrate the vertical component of force due to pressure: F L roof = 1,575 MN. 2) θ F=0 = 54,7 8.4 A simple reading gives F D = 6,9 N, W = 76 W. 8.5 Yes a reading gives ω = 83 rev/s. 191

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193 Fluid Mechanics Chapter 9 Introduction to compressible flow last edited April 2, Motivation Compressibility and its consequences Problem description The speed of pressure The measure of compressibility Thermodynamics of isentropic flow for a perfect gas Principle Pressure, temperature, density and the speed of sound Speed and cross-sectional area Isentropic flow in converging and diverging nozzles The perpendicular shock wave Compressible flow beyond frictionless pipe flow Exercises 207 These lecture notes are based on textbooks by White [13], Çengel & al.[16], and Munson & al.[18]. 9.1 Motivation In this chapter we investigate the fundamentals of compressible flows: those in which density varies significantly. This study should allow us to answer two questions: What happens in ideal (reversible) compressible flows? What is a shock wave, and how does it influence a flow? Video: pre-lecture briefing for this chapter by o.c. (CC-by) Compressibility and its consequences Problem description Ever since we have begun describing fluid flow in an extensive manner in chapter 4, we have restricted ourselves to incompressible flows those in which the fluid density ρ is uniform and constant. Now, ρ becomes one additional unknown property field. In some cases, solving for an additional unknown is not particularly challenging. For example, if we wish to calculate water temperature in a heat transfer case, or species concentration in a reacting flow, it is possible to first solve for the flow, and then solve for those particular property fields within the solution; thus the additional unknowns are solved separately with additional equations. The mathematical treatment of density, however, is much worse, because ρ appears in the momentum and mass conservation equations. Density directly affects the velocity fields, and so we have solve for ρ, p and V simultaneously. From a physical point of view, compressibility brings two new effects: 193

194 Density changes in fluids translate into large changes in fluid mechanical energy; these can be come very large in comparison to other forms of fluid energy, in particular kinetic energy. When ρ varies, fluids perform work upon themselves, accumulating and expending this energy, which translates into kinetic energy changes, and irreversible losses as heat transfer. When density is allowed to vary, pressure travels at finite speed within the fluid one could say that the fluid becomes appreciably squishy. The delay resulting from this finite pressure propagation speed becomes very significant if the fluid velocity is high, and it may even prevent the propagation of pressure waves in the upstream direction altogether. In this chapter, we wish to quantify both of those effects. By now, after eight chapters already where ρ was just an input constant, the reader should readily agree we have no hope of coming up with a general solution to compressible flow. Nevertheless, it is possible to reproduce and quantify the most important aspects of compressibility effects with relatively simple, one-dimensional air flow cases and this will suffice for us in this last exploratory chapter The speed of pressure We start by considering the speed of pressure. In compressible fluids, pressure changes propagate with a speed that increases with the magnitude of the pressure change. When this magnitude tends to zero (and the pressure wave reduces to that of a weak sound wave, with fully-reversible fluid compression and expansion) then the speed tends to a given speed which we name speed of sound and noted c. At ambient conditions, this speed is approximately km/h in air and roughly km/h in water. Let us, as a thought experiment, create a small pressure change inside a static fluid, and then travel along with the pressure wave. As we follow a wave which travels from right to left, we perceive fluid movement from left to right (fig. 9.1). We construct a small control volume encompassing and moving with the wave at the speed of sound c. Control volume fluid flow analysis does not frighten us anymore. From mass conservation (eq. 3/5 p.64) we write: ρca = (ρ + dρ)(c dv )A (9/1) 194 Figure 9.1 An infinitesimal pressure-difference wave is represented traveling from right to left inside a stationary fluid, in a one-dimensional tube. From the point of view of a control volume attached to (and traveling with) the wave, the flow is from left to right. Figure CC-0 o.c.

195 This equation 9/1 relates the density change across the wave to the velocity change between inlet and outlet. Re-arranging, and focusing only on the case where the pressure wave is extremely weak (as are sound waves in practice), we see the product dρ dv vanish and obtain: ρc = ρc ρ dv + c dρ dρ dv ρ dv = c dρ (9/2) In equation 9/2, which relates the speed of sound c to the density ρ, we would now like to eliminate the dv term. For this, we turn to the control volume momentum equation (eq: 3/8 p.65): F net = ṁ in V in + ṁ out V out pa (p + dp)a = ρcac + (ρ dρ)(c dv )A(c dv ) dpa = ρca [ c + (c dv )] dp = ρ dv (9/3) c We can now combine eqs. (9/2) and (9/3) to obtain: dp c = dρ This can be generalized for any pressure wave, and we state that p c = ρ s=cst. (9/4) (9/5) where the s = cst. subscript indicates that entropy s is constant. So now, we see that in a fluid the speed of sound the speed of pressure waves when they travel in a reversible (constant-entropy) manner is the square root of the derivative of pressure with respect to density. This allows us to compare it to another term, the bulk modulus of elasticity K, which represents the differential amount of pressure that one has to apply on a body to see its density increase by a certain percentage: K dp dρ ρ (9/6) where K is expressed in Pa. Inserting eq. 9/6 into eq. 9/5, we obtain a relationship which we had introduced already in chapter 6 (as eq. 6/20 p.138): K s=cst. c = (9/7) ρ With this equation 9/7 we see that in any fluid, the square of the speed of sound increases inversely proportionally to the density and proportionally to the modulus of elasticity (the hardness ) of the fluid. 195

196 9.2.3 The measure of compressibility The most useful measure of compressibility in fluid mechanics is the Mach number, which we already defined (with eq. 0/10 p.15) as: [Ma] V c (9/8) As we saw in chapter 6 ( p.138), [Ma] 2 is proportional to the ratio of net force to elastic force on fluid particles: at high Mach numbers the forces related to the compression and expansion of the fluid dominate its dynamics. Not all compressible flows feature high Mach numbers. In large-scale atmospheric weather for example, air flows mush slower than the speed of sound yet large density changes can occur because of the vertical (static) pressure gradient. The same is true in low-speed centrifugal machines because of centrifugal pressure gradients. In this chapter however, we choose to focus on the simplest of compressible flows: one-directional gas flows with Mach numbers approaching or exceeding Thermodynamics of isentropic flow for a perfect gas Principle When a gas is compressed or expanded, its pressure and temperature vary together with its density. Before we consider the dynamics of such phenomena, we need a robust model to relate those properties. This is the realm of thermodynamics. In short, from a macroscopic point of view, a gas behaves as a mechanical spring (fig. 9.2): one needs to provide it with energy as work in order to compress it, and one recovers energy as work when it expands. Three important points must be made regarding these compressions and expansions: 1. When gases are heated during the movement, the work transfers are increased; conversely they are decreased when the gas is cooled (fig. 9.3). Cases where no heat transfer occurs are called adiabatic; Figure 9.2 When they are compressed or expanded, the behavior of fluids can be modeled as if they were mechanical springs: they gain energy during compressions, and lose energy during expansions. These spring can be thought of as being fragile : when the movements are sudden, compressions involve more mechanical energy, and expansions involve less. Figure CC-by-sa Olivier Cleynen 196

197 Figure 9.3 Pressure-volume diagram of a perfect gas during a compression. When the fluid is heated up, more work is required; when it is cooled, less work is required. Opposite trends are observed during expansions. Figure CC-0 Olivier Cleynen Figure 9.4 When no heat transfer occurs, the a compression or expansion is said to be adiabatic. The temperature of fluids can still vary significantly because of the compression or expansion. When adiabatic evolutions are infinitely smooth, they are termed isentropic. Figure CC-0 Olivier Cleynen 197

198 2. The temperature of gases tends to increase when they are compressed, and tends to decrease when they expand, even when no heat transfer occurs (fig. 9.4); 3. A gas behaves as a fragile spring: the faster a compression occurs, and the more work is required to perform it. Conversely, the faster an expansion occurs, and the less work is recovered from it. Cases where evolutions occur infinitely slowly are used as a reference for comparison: such cases are termed reversible. Cases where the evolution is both adiabatic and reversible are named isentropic (iso-entropic, because they occur at constant entropy) Pressure, temperature, density and the speed of sound The quantitative relations between pressure, temperature and density of perfect gas during isentropic evolutions are classical results in the study of thermodynamics. In this section, we wish to re-write them with particular focus placed on the speed of sound and the Mach number. The goal is to arrive to eqs. 9/20 to 9/22 p.199, which we need in the upcoming sections to describe supersonic flow of gases. Let us then focus on the isentropic evolutions of a perfect gas. It is known from the study of thermodynamics that in isentropic flow from a point 1 to a point 2, the properties of a perfect gas are related one to another by the relations: ( ) ( ) γ 1 T1 v2 = (9/9) T 2 ( T1 T ( 2 p1 p 2 ) = ) = v 1 ( p1 p ( 2 v2 v 1 ) γ 1 γ (9/10) ) γ (9/11) for a perfect gas, during an isentropic (constant-entropy) evolution. It follows from eq. 9/11 that in an isentropic evolution, a perfect gas behaves such that pρ γ = k = cst., which enables us to re-write equation 9/5 for a perfect gas as: c = p ρ s=cst. = kγ ρ γ 1 = pρ γ γ ρ γ 1 = γpρ 1 The speed of sound in a perfect gas can therefore be expressed as: c = γrt (9/12) V [Ma] = (9/13) γrt for any evolution in a perfect gas. 198 Thus, in a perfect gas, the speed of sound depends only on the local temperature. We now introduce the concept of stagnation or total properties, a measure of the total amount of specific energy possessed by a fluid particle at a given

199 instant. They are the properties that the fluid would have if it was brought to rest in an isentropic manner: Using equation 9/13, we can re-write eq. (9/15) as: h 0 h V 2 (9/14) T 0 T c p 2 V 2 (9/15) h 0 = c p T 0 (9/16) T 0 T = c p T 2 V 2 = γr γ 1 T 2 V 2 ( γ 1 ) V 2 ( γ 1 ) = c 2 = 1 + [Ma] 2 2 Eqs. (9/9) to (9/11) can thus be re-written to express the transition from flow condition to stagnation condition: ( T0 ) ( γ 1 ) = 1 + [Ma] 2 (9/17) T 2 ( ) p0 [ ( γ 1 ) = 1 + [Ma] 2] γ γ 1 (9/18) p 2 ( ) ρ0 [ ( γ 1 ) = 1 + [Ma] 2] γ 1 1 (9/19) ρ 2 These three equations (9/17) to (9/19) express, for any point inside the flow of a perfect gas, the local Mach number [Ma] as a function of the ratio of the local (current) property (T, p or ρ) to the local total property (T 0, p 0, ρ 0 ). These ratios are always inferior or equal to one, by definition. Now, if now we define critical conditions as those that would occur if the fluid was exactly at [Ma] = 1, noting them with an asterisk, we can re-write eqs. (9/17) to (9/19) comparing stagnation properties to critical properties: ( ) T = 2 (9/20) T 0 γ + 1 ( p ) [ 2 = p 0 γ + 1 ( ρ ) [ 2 = γ + 1 ρ 0 ] γ γ 1 ] 1 γ 1 (9/21) (9/22) Here, we compare the local total property (e.g. the total temperature T 0 ) to the value the property needs to have for the flow to be at sonic speed (e.g. the critical temperature T ). This ratio does not depend at all on the fluid flow: it is merely a function of the (invariant) characteristics of the gas. This is it! It is important to understand that nothing in this section describes a particular flow movement. We have merely derived a convenient tool expressing flow properties as a function of their stagnation properties and the local Mach number, which we shall put to good use further down. 199

200 9.4 Speed and cross-sectional area Let us now go back to fluid dynamics. The last time we considered frictionless pipe flow, in chapter 5, we situation was rather simple. From the onedimensional mass conservation equation, eq. 5/1: ρ 1 V 1 A 1 = ρ 2 V 2 A 2 (9/23) it was possible to relate cross-sectional area A and velocity V easily. Now that ρ is also allowed to vary, the situation is decidedly more interesting. We wish to let static, pressurized air accelerate as fast as possible: letting V increase at a cost of a decease in ρ according to the rules of an isentropic, steady expansion. We know however, that according to mass conservation, ρv A has to remain constant all along the acceleration, so that only one given nozzle geometry (A = f (p,t )) will permit the desired acceleration. How can we find this geometry? We start from the conservation of mass inside our isentropic acceleration nozzle, and differentiating the terms: for any steady flow. ρav = cst. ln ρ + ln A + lnv = cst. dρ ρ + da A + dv V = 0 Let us set this equation 9/24 aside for a moment. During our isentropic expansion, the total energy of the fluid remains constant, thus the total specific enthalpy h 0 is constant, h + V 2 2 = cst. and here also we differentiate, obtaining: dh + V dv = 0 (9/24) In last relation, since the process is reversible and without external energy input, we can write the first term as dh = dp/ρ, yielding: for isentropic gas flow. 1 dp + V dv = 0 ρ 1 ρ = 1 V dv (9/25) dp This expression of 1/ρ is the relation we were after. Now, coming back to equation 9/24 and inserting eq. (9/25), we get: dρ da V dv + dp A + dv V = 0 [ dv 1 dρ ] V dp V 2 + da A = 0 200

201 In this last relation, since we are in an isentropic flow, we can insert an expression for the speed of sound, c 2 = p ρ derived from equation 9/5, obtaining: s dv [ 1 1 ] V c 2V 2 + da A = 0 dv [ 1 [Ma] 2 ] + da V A = 0 This leads us to the following equation relating area change and speed change in compressible gas flow: dv V = da A for isentropic, one-dimensional gas flow. 1 [Ma] 2 1 (9/26) This equation 9/26 is one of the most stunning equations of fluid dynamics. It relates the change in flow cross-section area, da, to the change in speed dv, for lossless flow. Let us examine how these two terms relate one to another, as displayed in figure 9.5: When [Ma] < 1, the term 1/([Ma] 2 1) is negative, and da and dv are always of opposite sign: an increase in area amounts to a decrease in speed, and a decrease in area amounts to an increase in speed. When [Ma] > 1, da and dv have the same sign, which perhaps disturbingly means that an increase in area amounts to an increase in speed, and a decrease in area amounts to a decrease in speed. Thus, if we want to let a static, compressed fluid accelerate as fast as possible, we need a converging nozzle up to [Ma] = 1, and a diverging nozzle onwards. 201

202 Figure 9.5 Changes in nozzle cross-section area have opposite effects when the Mach number [Ma] is lower or greater than 1. Note that because temperature changes as the gas compresses and expands, the local Mach number varies as the flow changes speed. It is thus possible for the flow to go from one regime to another during an expansion or compression. These effects are studied further down and described in figure 9.7 p.204. Figure CC-0 o.c. 9.5 Isentropic flow in converging and diverging nozzles 202 The change of between the subsonic and supersonic flow regimes, by definition, occurs at Mach one. When the flow is isentropic, equation 9/26 above tells us that the [Ma] = 1 case can only occur at a minimum area. Of course, there is no guarantee that a pipe flow will be able to become supersonic (fig. 9.6). For this, very large pressure ratios between inlet and outlet are required. Nevertheless, whichever the pressure ratio, according to eq. 9/26, a converging nozzle will only let an expanding fluid accelerate to Mach one (critical conditions). Changing the nozzle outlet area itself will not permit further acceleration unless an increase in nozzle area is generated beyond the critical point. This is a surprising behavior. In a converging-only nozzle, increasing the inlet pressure increases the outlet velocity only up until [Ma] = 1 is reached at the outlet. From there on, further increases in inlet pressure have no effect on the outlet pressure, which remains fixed at p outlet = p. The flow then reaches a maximum mass flow rate ṁ max. This condition is called choked flow. Once the flow is choked, adding a diverging section further down the nozzle may enable further acceleration, if the pressure ratio allows for it. Nevertheless, the pressure p T p throat at the throat remains constant at p T = p : from this point onwards, downstream conditions (in particular the pressure) are not propagated upstream anymore: the gas travels faster than the pressure waves inside it. With p T still equal to the critical pressure p, the mass flow through the nozzle cannot exceed ṁ max.

203 Figure 9.6 Non-dimensionalized properties of a gas undergoing an isentropic (i.e. shock-less) expansion through a converging-diverging nozzle. Figure CC-0 o.c. We can see, therefore, that it follows from equation 9/26 that there exists a maximum mass flow for any duct, which depends only on the stagnation properties of the fluid and the throat cross-sectional area. The mass flow, at any point in the duct, is quantified as: ṁ = ρca[ma] = p γrta[ma] RT ṁ = A[Ma]p 0 γ RT (γ 1)[Ma]2 2 γ 1 2(γ 1) (9/27) and ṁ will now reach a maximum when the flow is choked, i.e. reaches [Ma] = 1 at the throat, where its properties will be p = p T, T = T T, and the area A = A T : [ ṁ max = 2 γ + 1 ] γ +1 2(γ 1) A p 0 γ RT 0 (9/28) A general description of the properties of a perfect gas flowing through a converging-diverging duct is given in figs. 9.6 and

204 Figure 9.7 The pressure of a gas attempting an isentropic expansion through a converging-diverging duct, represented as a function of distance for different values of outlet pressure p b. The green curves show the reversible evolutions. As long as the back pressure p b is high enough, the flow is subsonic everywhere, and it remains reversible everywhere. When p T reaches p, the flow downstream of the throat becomes supersonic. In that case, if the back pressure is low enough, a fully-reversible expansion will occur. If not, then a shock wave will occur and the flow will return to subsonic regime downstream of the shock. Figure CC-0 o.c. 9.6 The perpendicular shock wave The movement of an object in a fluid at supersonic speeds (or a supersonic fluid flow over a stationary object, which produces the same effects) provokes pressure field changes that cannot propagate upstream. Thus, if a direction change or a pressure increase is required to flow past an obstacle, fluid particles will be subjected to very sudden changes. These localized changes are called shock waves. The design of bodies moving at supersonic speeds usually focuses on ensuring pressure and direction changes that are very gradual, and at the limit (e.g. far away from a supersonic aircraft) these are very small. The limiting case at the other side of the spectrum is that of the normal shock wave, the most loss-inducing (and simple!) type of shock wave. This will typically occur in front of a blunt supersonic body. A perpendicular shock wave (fig. 9.8) is studied with the following hypothesis: It is a constant energy process; 204 It it a one-dimensional process (i.e. the streamlines are all parallel to one another and perpendicular to the shock wave).

205 With these two hypothesis, we can compare the flow conditions upstream and downstream of a shock wave occurring in a tube using the continuity equation, which we wrote above as eq. 9/23: The energy equation in this case shrinks down to: Since, from equation 9/17 p.199, we have: T ( 01 γ 1 ) = 1 + [Ma] 2 1 T 1 2 T ( 02 γ 1 ) = 1 + [Ma] 2 2 T 2 2 ρ 1 A 1 V 1 = ρ 2 A 2 V 2 (9/29) T 01 = T 02 (9/30) and given that T 01 = T 02, dividing one by the other yields: γ 1 2 [Ma]2 1 T 2 = 1 + T γ 1 2 [Ma]2 2 (9/31) In a similar fashion, we can obtain: p 2 p 1 = 1 + γ [Ma] γ [Ma] 2 2 (9/32) Let us set aside these two equations 9/31 and 9/32 and compare conditions upstream and downstream starting with mass conservation, and re-writing it as a function of the Mach number: ρ 1 V 1 = ρ 2 V 2 ρ 1 = V 2 = [Ma] 2a 2 ρ 2 V 1 [Ma] 1 a 1 = [Ma] 2 γrt2 [Ma] 1 γrt1 [Ma] 1 = p ( ) 1 2 T1 2 (9/33) [Ma] 2 p 1 T 2 So far, we have not obtained anything spectacular, but now, inserting both equations 9/31 and 9/32 into this last equation 9/33, we obtain the staggering Figure 9.8 A normal shock wave occurring in a constant-section duct. All properties change except for the total temperature T 0, which, since the transformation is at constant energy, remain constant. Figure CC-0 o.c. 205

206 result: [Ma] 1 [Ma] 2 = 1 + γ [Ma] γ [Ma] γ 1 2 [Ma] γ 1 2 [Ma] (9/34) This equation is not interesting because of its mathematical properties, but because of its contents: the ratio [Ma] 1 /[Ma] 2 depends only on [Ma] 1 and [Ma] 2! In other words, if the shock wave is perpendicular, there is only one possible post-shock Mach number for any given upstream Mach number. After some algebra, [Ma] 2 can be expressed as [Ma] 2 2 = [Ma] γ 1 γ 2[Ma] 2 1 γ 1 1 (9/35) Values for [Ma] 2 as a function of [Ma] 1 according to equation 9/35 for particular values of γ can be tabulated or recorded graphically, allowing us to predict properties of the flow across a shock wave. We can therefore quantify the pressure and temperature increases, and velocity decrease, that occur though a perpendicular shock wave. 9.7 Compressible flow beyond frictionless pipe flow In this chapter, we have restricted ourselves to one-directional steady flow, i.e. flows inside a pipe; within these flows, we have neglected all losses but those which occur in perpendicular shock waves. This is enough to describe and understand the most important phenomena in compressible flow. Of course, in reality most compressible flows are much more complex than that. Like other flows, compressible flows have boundary layers, feature laminarity and turbulence, heat transfer, unsteadiness, and more. Shock waves occur not just perpendicularly to flow direction, but also in oblique patterns, reflecting against walls and interacting with boundary layers. Temperature variations in compressible flow are often high enough to provoke phase changes, which complicates the calculation of thermodynamic properties as well as flow dynamics and may result in spectacular visual patterns. All those phenomena are beyond the scope of this study, but the reader can hopefully climb up to more advanced topics such as high-speed aerodynamics, or rocket engine and turbomachine design, where they are prominently featured and can surely make for some exciting learning opportunities. Video: watch a 140-thousandton rocket lift away humans on their first ever trip to the moon. Compressible flow of expanding exhaust gases is what allows the maximum amount of thrust to be obtained from a given amount of combustion energy in rocket engines. by NASA (public domain) 206

207 Exercise sheet 9 (Compressible flow) last edited April 2, 2018 These lecture notes are based on textbooks by White [13], Çengel & al.[16], and Munson & al.[18]. Except otherwise indicated, we assume that fluids are Newtonian, and that: ρ water = kg m 3 ; p atm. = 1 bar; ρ atm. = 1,225 kg m 3 ; T atm. = 11,3 C; µ atm. = 1, N s m 2 ; д = 9,81 m s 2. Air is modeled as a perfect gas (R air = 287 J K 1 kg 1 ; γ air = 1,4; c pair = J kg 1 K 1 ). The speed of sound c in a perfect gas is a local property expressed as: c = γrt (9/12) In isentropic, one-dimensional flow, we accept that the mass flow ṁ is quantified as: ṁ = A[Ma]p 0 γ RT (γ 1)[Ma]2 2 γ 1 2(γ 1) This expression reaches a maximum ṁ max when the flow is choked: [ ṁ max = 2 γ + 1 (9/27) ] γ +1 2(γ 1) A p 0 γ RT 0 (9/28) The properties of air flowing through a converging-diverging nozzle are described in fig. 9.9, and fig describes air property changes through a perpendicular shock wave. 207

208 208 Figure 9.9 Properties of air (modeled as a perfect gas) as it expands through a convergingdiverging nozzle. These are numerical values for equations (9/17) to (9/19). In this figure, the perfect gas parameter γ is noted k. Data also includes the parameter [Ma] V /c (speed non-dimensionalized relative to the speed of sound at the throat). Figure Çengel & Cimbala 2010 [16]

209 Figure 9.10 Properties of air (modeled as a perfect gas) as it passes through a perpendicular shock wave. These are numerical values calculated based on equation 9/35. In this figure the perfect gas parameter γ is noted k. Figure Çengel & Cimbala 2010 [16] 209

210 9.1 Flow in a converging nozzle Çengel & al. [16] E12-5 Air from a 4 mm hole in a car tire leaks into the atmosphere. The pressure and temperature in the tire are 314 kpa and 25 C. The exterior air conditions are 94 kpa and 16 C. If the flow can be approximated as isentropic, what is the air flow rate? 9.2 Flow in a converging-diverging nozzle Exam ws , see also White [13] E9.4 An experiment is conducted where compressed air from a large reservoir is expanded and accelerated through a nozzle (figure 9.11). There is no external work transfer to or from the air, and the expansion can satisfactorily be modeled as a one-dimensional isentropic process (isentropic meaning that the process is fully reversible, i.e. without losses to friction, and adiabatic, i.e. without heat transfer). Figure 9.11 A nozzle in which compressed air in a reservoir (left) is expanded as it accelerates towards the right. The back pressure p b is controlled by a valve (represented on the right). The flow conditions of interest in this exercise are labeled 1, 2 and 3. The air (c pair = J kg 1 K 1 ) flows through a converging-diverging nozzle. The entrance conditions into the nozzle are A 1 = 0,15 m 2, p 1 = 620 kpa, V 1 = 180 m s 1, and T 1 = 430 K. In the expanding section of the nozzle, we wish to know the conditions at a point 2 of cross-sectional area A 2 = A 1 = 0,15 m What is the mass flow through the nozzle? 2. If we wish the flow to become sonic (i.e. reach [Ma] = 1), what is approximately the required throat area A T? 3. Which two flow regimes can be generated at section 2? What is approximately the pressure p 2 required for each? 210 The nozzle is now operated so that the flow becomes sonic at the throat, and accelerates to supersonic speeds in the expanding section of the nozzle. The nozzle cross-sectional area increases further downstream of point 2, up until it reaches an area A 3 = 0,33 m 2. Downstream of point 3, the duct is of constant area and the back pressure p b is set to p b = p 3 = p 2.

211 4. Describe qualitatively (i.e. without numerical data) the pressure distribution throughout the nozzle. Finally, we wish to expand the air with a fully-isentropic process, expanding the air from the reservoir to point 3 without a shock wave. 5. Propose and quantify one modification to the experimental set-up that would allow an isentropic, supersonic expansion of the flow into the outlet of area A Capsule re-entry CC-0 o.c. A Soyuz capsule incoming from the international space station is passing through Mach 3,5 at a pressure of 0,2 bar and temperature of 50 C. A bow shock wave is generated in front of the bottom surface of the capsule, which is approximately normal to the flow for the particles directly in the path of the capsule. What are the conditions (temperature, velocity, density and pressure) in between the shock wave and the capsule shield? 9.4 Explosion shock wave CC-0 o.c. In an unfortunate coincidence, on the morning following the fluid dynamics exam, the red Tesla S of the lecturer explodes on a parking lot (fortunately, no one was hurt). The explosion creates an spherical shock wave propagating radially into still air. At a given moment, the pressure inside the spherical bubble is kpa. The ambient atmospheric conditions are 17 C. At what speed does the shock wave propagate, and what is the air speed immediately behind the wave? 9.5 Home-made supersonic tunnel CC-0 o.c. A PhD student is thinking about constructing his/her own supersonic wind tunnel. To this purpose, s/he designs a converging-diverging tube with variable square crosssection. The throat T is 0,1 0,1 m; the tunnel expands progressively to an outlet nozzle of 0,13 0,13 m cross-section. The outlet exits directly in the ambient atmosphere. On the inlet side of the tunnel, the student hopes to install a very large concrete reservoir (fig. 9.12). 1. What is the maximum reservoir pressure for which the flow in the tunnel is subsonic everywhere? 2. What reservoir pressure would be required to generate a shock-less expansion to supersonic flow in the tunnel? 3. If the outlet temperature was to be ambient (11,3 C), what would the outlet velocity be? 4. Which mass flow rate would then be passing in the tunnel? 5. Describe qualitatively (i.e. without numerical data) the pressure distribution throughout the nozzle for both flow cases above. 211

212 Figure 9.12 A nozzle in which compressed air in a reservoir (left) is expanded as it accelerates towards the right. The back (outlet) pressure p b is always atmospheric. In order to impress his/her friends, the PhD student wishes to generate a perpendicular shock wave inside the tunnel at exactly one desired position (labeled 1 in fig above). At this position, the cross-section area is 0,01439 m Describe qualitatively (i.e. without numerical data) the pressure distribution throughout the nozzle for this case, along the two other pressure distributions. 7. [difficult question] What reservoir pressure would be required to generate the shock wave at exactly the desired position? 8. What would the outlet velocity approximately be? 212